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.

From E=W_0​+(1/2)​mv_max²​⇒v_max​= √[(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

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.

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.

• 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)
   

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=6.6×10^34​/9.95×10⁻²⁵
0λ=0.66×10⁻⁹
λ=6.6×10⁻¹⁰m

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.

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

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.

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.

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)

 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.

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.

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

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.

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
 

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.

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.