Physics: Topic-wise Test- 6 - NEET MCQ

# Physics: Topic-wise Test- 6 - NEET MCQ

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## 30 Questions MCQ Test NEET Mock Test Series 2025 - Physics: Topic-wise Test- 6

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*Multiple options can be correct
Physics: Topic-wise Test- 6 - Question 1

### If radiation of allow wavelengths from ultraviolet to infrared is passed through hydrogen agas at room temperature, absorption lines will be observed in the :

Detailed Solution for Physics: Topic-wise Test- 6 - Question 1

At room temperature, nearly all the atoms in hydrogen gas will be in the ground state. When light passes through the gas, photons are absorbed, causing electrons to make transitions to higher states and creating absorption lines. These lines correspond to the Lyman series since that is the series of transitions involving the ground state or n = 1 level. Since there are virtually no atoms in higher energy states, photons corresponding to transitions from n > 2 to higher states will not be absorbed.

Physics: Topic-wise Test- 6 - Question 2

### In a photo-emissive cell, with exciting wavelength l, the maximum kinetic energy of electron is K. If the exciting wavelength is changed to the kinetic energy of the fastest emitted electron will be :

Detailed Solution for Physics: Topic-wise Test- 6 - Question 2

From E=W0​+(1/2)​mvmax2​⇒vmax​= √[(2E/m)​−(2W0/m)]​​​
where E= hc​/λ
If wavelength of incident light charges from λ to 3λ/4​ (Decreases)
Let energy of incident light charges from E and speed of fastest electron changes from v to v′ then
v′=√[(2E′/m)​−(2W0/m)​​​]
As E∝1/λ​⇒E′(4/3)​E
Hence v′=√[(2(4/3​)E/m)​−(2W0/m)​​]​
⇒v′=(4/3)1/2 [(2E/m)​− (2W0/m(4/3)1/2​)]​​
So, v′>(4/3)1/2v

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Physics: Topic-wise Test- 6 - Question 3

### The frequency and the intensity of a beam of light falling on the surface of photoelectric material are increased by a factor of two (Treating efficiency of photoelectron generation as constant). This will :

Detailed Solution for Physics: Topic-wise Test- 6 - Question 3

Physics: Topic-wise Test- 6 - Question 4

Light coming from a discharge tube filled with hydrogen falls on the cathode of the photoelectric cell. The work function of the surface of cathode is 4eV. Which one of the following values of the anode voltage (in Volts) with respect to the cathode will likely to make the photo current zero.

Detailed Solution for Physics: Topic-wise Test- 6 - Question 4

Given,
Work function, ϕ=4eV
for Hydrogen atom, it can provide a max of 13.6 eV of energy by exciting its electron.
∴ e(Stopping potential) ≥ 13.6 - 4
≥ 9.6
i.e. V = 10 V will make photo current zero.

Physics: Topic-wise Test- 6 - Question 5

In a photoelectric experiment, the potential difference V that must be maintained between the illuminated surface and the collector so as just to prevent any electron from reaching the collector is determined for different frequencies f of the incident illumination. The graph obtained is shown. The maximum kinetic energy of the electrons emitted at frequency f1 is

Detailed Solution for Physics: Topic-wise Test- 6 - Question 5

*Multiple options can be correct
Physics: Topic-wise Test- 6 - Question 6

In photoelectric effect, stopping potential depends on

Detailed Solution for Physics: Topic-wise Test- 6 - Question 6

The stopping potential depends on frequency of incident light and nature of the emitter material.for a given frequency of incident light, it is independent of its intensity.

Physics: Topic-wise Test- 6 - Question 7

Two electrons are moving with the same speed v. One electron enters a region of uniform electric field while the other enters a region of uniform magnetic field, then after sometime if the de-Broglie wavelengths of the two are l1 and l2 then :

Physics: Topic-wise Test- 6 - Question 8

An electron in hydrogen atom first jumps from second excited state to first excited state and then, from first excited state to ground state. Let the ratio of wavelength, momentum and energy of photons in the two cases by x, y and z, then select the wrong answers :

Detailed Solution for Physics: Topic-wise Test- 6 - Question 8
• 1/λ=R[(1/nf2)-(1/nl2)
• Therefore, for 3->2 , we get,
• 1/λ1=R[(1/22)-(1/32)]=R[(1/4)-(1/9)]=5R/36
• Therefore, λ1=36/5R
• Similarly, for 2->1, we get,
• 1/λ2=R[(1/12)-(1/22)]=R[(1-(1/4)]=3R/4
• Therefore, λ2=4/3R
• Therefore, the ratio is,
• x=λ12=(36/5R) x (3R/4)=27/5
• hence, option B is wrong.
•
• Now, the momentum is given as,
• p=h/λ
• Therefore, 3->2
• p1=h/λ1=5Rh/36
• similarly, for 2->1, we get,
• p2 =h/λ2=3Rh/4
• therefore, the ratio is,
• y=p1/p2=(5Rh/36)x(4/3Rh)=5/27
• hence, option C is correct.
• Now the energy difference between the two levels is,
• ΔE=En-1-En
• Therefore, for 3->2
• ΔE1=E2-E3=(E1/22)-(E1/32)=E1[(1/4)-(1/9)]=5E1/36
• Similarly, 2->1
• ΔE2=E1-E2=E1-(E1/22)=E1[1-(1/4)]=3E1/4
• Therefore, the ratio is,
• z= ΔE1/ΔE2=(5E1/36) X (4/3E1)=5/27
• hence, option A and D are both correct.
• therefore, we can say that the wrong is option (B)

Physics: Topic-wise Test- 6 - Question 9

An electron is in an excited state in hydrogen-like atom. It has a total energy of –3.4 eV. If the kinetic energy of the electron is E and its de-Broglie wavelength is l, then

Detailed Solution for Physics: Topic-wise Test- 6 - Question 9

The potential energy of an electron is twice the kinetic energy(with negative sign) of an electron.
PE=2E
The total energy is: TE=PE+KE
−3.4=−2×3.4+KE
KE=3.4eV
Let p be the momentum of an electron and m be the mass of an electron.
E=p2​/2m
p=√2​mE

Now, the De-Broglie wavelength associated with an electron is:
λ=h/p​
λ=h/√2​mE​
λ=6.6×1034​/√2​×9.1×10−31×(−3.4)×1.6×10−19
λ=6.6×1034​/9.95×10−25
0λ=0.66×10−9
λ=6.6×10−10m

*Multiple options can be correct
Physics: Topic-wise Test- 6 - Question 10

A particular hydrogen like atom has its ground state binding "energy 122.4 eV. Its is in ground state. Then :

Detailed Solution for Physics: Topic-wise Test- 6 - Question 10

En​= −z213.6eV/ n2
for ground state, n=1
−122.4=−Z213.6eV
When a proton interacts that
EP​=hV=E1​−E0​=Z213.6eV[(1/12​)−(1/22)​]
=9×13.6eV [3/4]
=91.8eV
So the Atomic number is 3 , a photon  of 91.8eV and an electron of 8.2eV are emitted when 100eV electron interacts e− interact with this atom.

*Multiple options can be correct
Physics: Topic-wise Test- 6 - Question 11

The electron in a hydrogen atom makes a transition n1 ¾→ n2, where n1 & n2 are the principal quantum numbers of the two states. Assume the Bohr model to be valid. The time period of the electron in the initial state is eight times that in the final state. The possible values of n1 & n2 are :

Detailed Solution for Physics: Topic-wise Test- 6 - Question 11

For an electron revolving in nth orbit around the nucleus of hydrogen atom, Fcentrifugal​=Fcoulomb​
Thus,  mvn2/ rn​​​=​e2​/4πϵo​rn2
Also, mvn​rn​=nh/2π​
On solving these two we get: vn​= e21/2ϵo​hn ​and rn​=( ϵo​h2/πme2 ​) n2
Time period, T=2πrn/vn​​​⟹T∝n3
Thus, 8T/T​=(​n1/n2)3⟹n1​=2n2
Thus, n1​=4,n2​=2  and   n1​=6,n2​=3

Physics: Topic-wise Test- 6 - Question 12

In hydrogen atom the kinetic energy of electron in an orbit of radius r is given by

Detailed Solution for Physics: Topic-wise Test- 6 - Question 12

K.E. of nth orbit
=> (1/k) Ze2/2r
For H atom,
K.E.=(1/4πε) x (e2/2r)

Physics: Topic-wise Test- 6 - Question 13

According to Bohr model of hydrogen atom, the radius of stationary orbit characterized by the principal quantum number n is proportional to​

Detailed Solution for Physics: Topic-wise Test- 6 - Question 13

r=(0.529Å)n2/7
r ∝n2

Physics: Topic-wise Test- 6 - Question 14

Select an incorrect alternative:
i. the radius of the nth orbit is proprtional to n2
ii. the total energy of the electron in the nth orbit is inversely proportional to n
iii. the angular momentum of the electron in nth orbit is an integral multiple of h/2π
iv. the magnitude of potential energy of the electron in any orbit is greater than its kinetic energy​

Detailed Solution for Physics: Topic-wise Test- 6 - Question 14

Statement i. Radius of Bohr's orbit of hydrogen atom is given by
r= n2h2​/4π2mKze2
or, r=(0.59A˚)(n2​/z)
So, from expression we found r∝n2
Hence the 1st statement is correct.
Statement ii.

We know that
En=-13.6 x z2/n2
So, En ∝1/n2
Hence the 2nd statement is wrong.
Statement iii.
Bohr defined these stable orbits in his second postulate. According to this postulate:

• An electron revolves around the nucleus in orbits
• The angular momentum of revolution is an integral multiple of h/2π – where Planck’s constant [h = 6.6 x 10-34 J-s].
• Hence, the angular momentum (L) of the orbiting electron is: L = nh/2 π

Hence the 3rd statement is correct.
Statement iv.
According to Bohr's theory
Angular momentum of electron in an orbit will be Integral multiple of (h/2π)
Magnitude of potential energy is twice of kinetic energy of electron in an orbit
∣P.E∣=2∣K.E∣
K.E=(13.6ev)( z2/n2)​
Hence, The 4th statement is correct.

Physics: Topic-wise Test- 6 - Question 15

Which of these is true?

Detailed Solution for Physics: Topic-wise Test- 6 - Question 15

Rutherford declared that the "nucleus" (as he now called it) was indeed positively charged, based on the result of experiments exploring the scattering of alpha particles in various gases.

Physics: Topic-wise Test- 6 - Question 16

Which of the following is/are deduced from the Rutherford’s scattering experiment?
(1) There are neutrons inside the nucleus.
(2) The sign of the charge of the nuclei is the same as the sign of alpha particles.
(3) Electrons are embedded in the nucleus.​

Detailed Solution for Physics: Topic-wise Test- 6 - Question 16

When he sent alpha particles during his gold foil experiment he observed that most of the rays moved away from the nucleus.As we know that like charges always repel ,he concluded that the charge the nuclei is same as the charge of alpha particles

Physics: Topic-wise Test- 6 - Question 17

In scattering, the impact parameter b is defined as the:

Detailed Solution for Physics: Topic-wise Test- 6 - Question 17

The impact parameter is defined as the perpendicular distance between the path of a projectile and the center of a potential field created by an object that the projectile is approaching. It is often referred to in nuclear physics (see Rutherford scattering) and in classical mechanics.

Physics: Topic-wise Test- 6 - Question 18

The alpha particle scattering experiment was carried out by:

Detailed Solution for Physics: Topic-wise Test- 6 - Question 18

Rutherford Alpha Particle Scattering Experiment.Rutherford directed beams of alpha particles (which are the nuclei of helium atoms and hence positively charged) at thin gold foil to test this model and noted how the alpha particles scattered from the foil.

Physics: Topic-wise Test- 6 - Question 19

We know that the Rutherford model of the atom is superior to the Thompson model because when alpha particles are scattered from atoms:

Detailed Solution for Physics: Topic-wise Test- 6 - Question 19

In ruther ford experiment he suggest that all the positive charge and mass are concentrated at the centre when he bombarded the alpha partical which is dipositive in nature and when it is more close to centre it get deflect to a large angle and with increase of closenes to centre its deflection angle increase and some alpha partical deflect to 180 degree so it prove that all the positive charge and mass are concentrated at the centre where as acccording to thomson atom is hard solid sphere in which its total +ve charge and mass uniformalyy distributed on the surface and electrone reside as seed in watermelon ( plum pudding model)

Physics: Topic-wise Test- 6 - Question 20

In Rutherford’s experiment, a thin gold foil was bombarded with alpha particles. According to Thomson’s “plum-pudding” model of the atom, what should have happened?

Detailed Solution for Physics: Topic-wise Test- 6 - Question 20

In Rutherford's experiment, a thin gold foil was bombarded with alpha particles. According to Thomson's "plum-pudding" where atoms are to be made up of electrons embedded in positive charge cloud, Alpha particles should have passed through the foil with little or no deflection because the diffused, massless positive charge cloud will not pose any obstacle to the fast moving heavy alpha particle beam.

Physics: Topic-wise Test- 6 - Question 21

The distance of closest approach when a 15.0 MeV proton approaches gold nucleus (Z = 79) is​

Detailed Solution for Physics: Topic-wise Test- 6 - Question 21

Explanation : E = 15.0MeV

= 15 * 106 eV

= 15 * 106 * 1.6 * 10-19 J

= 15 * 1.6 * 10-13 J

E = (1/4πεo)*(ze2/r02)

r0 = (1/4πεo)*(ze2/E)

Physics: Topic-wise Test- 6 - Question 22

What percentage of the mass of an atom is concentrated in the nucleus?

Detailed Solution for Physics: Topic-wise Test- 6 - Question 22

More than 99.99% of the mass of any atom is concentrated in its nucleus. If the mass of protons and neutrons (which are in the nucleus of every atom) is approximately one atomic mass unit, then the relative mass of an electron is 0.0005 atomic mass units.

Physics: Topic-wise Test- 6 - Question 23

Plutonium decays with a half-life of 24000 years. If the plutonium is stored for 72000 years, then the fraction of plutonium that remains is

Detailed Solution for Physics: Topic-wise Test- 6 - Question 23

t1/2​=24000 yrs
It is stored for 3 half-lives.
Fraction remaining is (1/2​)3=1/8

Physics: Topic-wise Test- 6 - Question 24

In a nuclear reaction which of the following is conserved

Physics: Topic-wise Test- 6 - Question 25

In what units is mass measured on the atomic scale?

Detailed Solution for Physics: Topic-wise Test- 6 - Question 25

The atomic mass of an element is the average mass of the atoms of an element measured in atomic mass unit (amu, also known as daltons, D). The atomic mass is a weighted average of all of the isotopes of that element, in which the mass of each isotope is multiplied by the abundance of that particular isotope.

Physics: Topic-wise Test- 6 - Question 26

If 10% of a substance decays in 10 days, then approximate percentage of substance left after 24 days is

Detailed Solution for Physics: Topic-wise Test- 6 - Question 26

Given, 10% of substance decays in 10 days
So, let [A]o=100, [A]t=100-10=90
So, K=(2.303/t) log ([A]o/[A]t)
K= (2.303/10) log (100/90)
K= (2.303/10) (log10-log9)
K= (2.303/10)91-0.9542)     [log9=0.9542]
K= (2.303/10) x 0.0458 day-1
So, after 24 days,
t= (2.303/K) log (100/[A]t)
24=2.303x10/2.303x0.0458) log [100/[A]t]
log[100/[A]t]=(24x0.0458)/10=0.1099
100/[A]t=antiog(0.1099)=1.288
So, [A]t=100/1.288≈100/1.3
[A]t=76.9≈77%
The closest answer is 78% which is option A. So the correct answer would be option A

Physics: Topic-wise Test- 6 - Question 27

Choose the WRONG statement. A thermonuclear fusion reactor is better than a fission reactor for the following reason:

Physics: Topic-wise Test- 6 - Question 28

Which of the following is true for a forward-biased diode?

Detailed Solution for Physics: Topic-wise Test- 6 - Question 28

There are two types of ions -anions are the - ve ones & cations are the +ve ones.As we know current flow from +ve to - ve, so During electrolysis or forward biasing anions move to anodes that are +ve in nature whereas cations move to cathodes which are -ve in nature.

Physics: Topic-wise Test- 6 - Question 29

The diode current depends on which of the following:

Detailed Solution for Physics: Topic-wise Test- 6 - Question 29

The diode will reach its trigger point at lower voltage with rise in ambient temp, and at higher voltage for lower temp. As increased in temp, it accounts larger diode current than its previous current. When we talk about REVERSE SATURATION CURRENT, yes the diode current is dependable on it, because the (IS) i.e., reverse saturation current mainly depends on temp.

Physics: Topic-wise Test- 6 - Question 30

In an unbiased p-n junction, holes diffuse from the p-region to n-region because

Detailed Solution for Physics: Topic-wise Test- 6 - Question 30

The diffusion of charge carriers across a junction takes place from the region of higher concentration to the region of lower concentration. In this case, the p-region has greater concentration of holes than the n-region. Hence, in an unbiased p-n junction, holes diffuse from the p-region to the n-region.

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