Physics: Topic-wise Test- 6 - NEET MCQ

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.

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.

If light of certain frequency, greater than the threshold frequency, is incident of the metal surface the photoelectric current obtained is directly proportional to the number of photo-electrons emitted. On increasing the frequency of incident light (or) by increasing the anode potential the speed of photo-electrons emitted changes but they have no effect on the number of electrons emitted. But by increasing the intensity of the incident light more photo-electrons will be emitted and thereby increasing the photoelectric current. 
Also, if we increase the area on which the light is incident we can get more electrons emitted which increases the photoelectric current.
 

• 1/λ=R[(1/n_f²)-(1/n_l²)
• 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=λ₁/λ₂=(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=E_n-1-E_n
• 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)
   

E_n​= −z²13.6eV/ n²​
for ground state, n=1
−122.4=−Z²13.6eV
When a proton interacts that 
E_P​=hV=E_1​−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.

For an electron revolving in nth orbit around the nucleus of hydrogen atom, F_centrifugal​=F_coulomb​
Thus,  mv_n²/ r_n​​​=​e²​/4πϵ_o​r_n²
Also, mv_n​r_n​=nh/2π​
On solving these two we get: v_n​= e²1/2ϵ_o​hn ​and r_n​=( ϵ_o​h²/πme² ​) n² 
Time period, T=2πr_n/v_n​​​⟹T∝n³
Thus, 8T/T​=(​n₁/n₂)³⟹n₁​=2n₂​
Thus, n₁​=4,n_2​=2  and   n₁​=6,n₂​=3 

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

r=(0.529Å)n²/7
r ∝n²

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.

Bohr used conservation of angular momentum.
For stationary orbits, Angular momentum Iω=nh2π
where n=1,2,3,...etc

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.

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

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.

Correct Answer :- b

Explanation : E = 15.0MeV

= 15 * 10⁶ eV

= 15 * 10⁶ * 1.6 * 10^-19 J

= 15 * 1.6 * 10^-13 J

E = (1/4πε_o)*(ze²/r₀²)

r₀ = (1/4πε_o)*(ze²/E)

Physicist Ernest Rutherford established the nuclear theory of the atom with his gold-foil experiment. When he shot a beam of alpha particles at a sheet of gold foil, a few of the particles were deflected. He concluded that a tiny, dense nucleus was causing the deflections.

Most of the α-particle passed through the foil straight without suffering any change in their direction. This shows that most of the space inside the atom is empty or hollow. 
A small fraction of α-particles was deflected through small angles and a few through larger angles. For this to happen α- particles (positively charged) must approach a heavy positively charged core inside the atom (like charges repel each other). This heavy positively charged core inside the atom was named as the nucleus.

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.

t_1/2​=24000 yrs
It is stored for 3 half-lives.
Fraction remaining is (1/2​)³=1/8

In nuclear reactors, fission of uranium is carried out by bombarding  a slow moving neutron on uranium which results in emission of 3 more neutrons along with the byproducts  So, cadmium rods are used to absorb extra neutrons in order to achieve controlled chain reaction.

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.

Half-life of Bi²¹⁰=5 days
∴k= 0.693/(t1/2) ​=(0.693/5) ​day⁻¹
Using k=(2.303/t)​ log (a/a-x)
(where a = a_0​, (let) ⇒x=7/8 ​a₀​, t is time taken in decay and k is rate constant)
We get, t=(2.303×5/0.693)log a₀​/(1/8)a₀​​
= (2.303×5/0.693) ​log8=15days
 

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
 

Neutron is unstable outside the nucleus.
It's decaying equation is given below.
0​ⁿ¹⇒¹₁​H+⁰₋₁e+v

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.

The varactor diode is used in a place where the variable capacitance is required, and that capacitance is controlled with the help of the voltage. The Varactor diode is also known as the Varicap, Voltcap, Voltage variable capacitance or Tunning diode.

In a p-type semiconductor, the holes are the majority carriers, while the electrons are the minority carriers. A p-type semiconductor is obtained when trivalent atoms, such as aluminium, are doped in silicon atoms.

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.