Test: Dual Nature of Radiation & Matter - 2 - JEE MCQ

Since, P_r​=P_b​ 
r for red and b for blue.
P_r​=P_b​
or, n_r​× (hc/λr)  ​=nb​× (hc​/λ_b)​
or,  ​(n_r/n_b)​​= λ_r​​/λ_b ​
Since, the wavelength of red bulb is greater than the wavelength of blue bulb.
or,  n_r​>n_b

Current is 0.16×10⁻⁶ Amp it means 0.16×10−6 Coulomb charge is flowing per second
So, n=0.16×10−6C /1.6×10⁻¹⁹C ​=10¹² electrons are generated per second
Now we notice that one photon has energy E, E=hc/λ=​=(6.62×10⁻³⁴Js×3×10⁸ms^-1)/(5000×10−10m) ​=3.972×10⁻¹⁹Joule
So, number of photon in 10−3W will be N=10⁻³/3.972×10⁻¹⁹ ​=0.25×10¹⁶ this is number of photons incident per second
So required percentage is (n/N)×100=10¹⁴/(0.25×10¹⁶) ​=0.04%

The maximum kinetic energy for the photoelectrons is 
E_max​=hν−ϕ
where, ν is the frequency of incident light and ϕ is photoelectric work function of metal.
eV_0​=hν−ϕ ...................(1)
where, V₀​ is the stopping potential and e is the electronic charge.
When, the frequency of light in a photoelectric experiment is doubled, 
eV₀′​=2hν−ϕ 
eV₀′​=2[hν−(ϕ/2​)].........................(2)
From the above two equations we can say that the K.E. in (2) is more than double of K.E in (1). Hence, when the frequency of light in a photoelectric experiment is doubled, the stopping potential becomes more than double.
So, the answer is option (C).
 

As we know that the stopping potential of the photoelectron is equal to the maximum kinetic energy of the photoelectron,
KEmax​=10.4V
Now, in photoelectric effect,
Energy of incident radiation (E_in​) = work function + K.Emax​
⇒ E_in​=1.7+10.4
⇒ E_in​=12.1eV
Now, for 0 hydrogen atom,
Energy of first energy level, E₁​=−13.6eV
Energy of second energy level, E₂​=−3.4eV
Energy of third energy level, E₃​=−1.5eV
Hence, a transition from third to first energy level will result in emission of radiation of energy = E₃​−E₁​=12.1eV which is same as the energy of incident radiation of above photoelectric effect.
Thus, correct answer is n=3 to 1
 

When photon strikes with the electron it completely transfers it’s energy to the electron as during photoelectric experiment the threshold frequency required is used by the electron to eject from the atom which is also called the work function and the remaining energy in electron is kinetic energy which can be measured. Now (before collision) since photon is a particle it must have mass and thus it has energy equivalent to E=mc2 .
After collision when it completely transfers it’s energy to electron thus E=0
And therefore 0=mc2 thus i guess photon vanishes.
 

When the source is moved away from the emitter, intensity of the incident radiation decreases but frequency remains the same so there will be no change in the stopping potential. Thus, it remains constant.

Saturation current is the maximum current possible and it will be directly proportional to the number of number of electrons falling on collector plate per second which depend on number of photons
 incident on the cathode as one photon contribute in one electron and the number of photons is actually
 proportional to intensity which varies 
As intensity I∝1/r2​, where r is the distance 
So the correct graph will be decreasing with power2 of distance and it will be rapidly decreasing with a higher value of r.
 

Given, the maximum kinetic energy: K_max​=4eV
If V₀​ be the stopping potential, then K_max​=eV_0​
⇒eV_0​=4eV 
⇒V₀​=4V

As,
Maximum K.E= incident photon energy − work function
(½)​mV₁²​=2W−W=W⟶(1)
and
V_e​mV₂²​=5W−W=4W⟶(2)
from (1) and (2)
V₁/V₂​ ​​=1/2​
V_1​:V₂​=1:2

We know that,
eV=(hc/λ)-w
V=(hc/eλ)-(w/e)
Arithmetic progression =>V₂=(V₁+V₂)/2
Now,
(hc/eλ₂)-w/e=1/2[(hc/eλ₁)-(w/e) +(hc/eλ₃) -(w/e)]
=>1/ λ₂=1/2[(1/ λ₁)+(1/λ₃)]
=>2/ λ₂=1/ λ₁ + 1/λ₃
Hence the correct answer is harmonic Progression.

When 5eV is incident the kinetic energy is 2eV it simply means the work function is W=5eV−2eV=3eV
Similarly, when 6eV is incident the kinetic energy should be   6eV−W=6eV−3eV=3eV
 it simply means to stop them we need a negative potential at anode equal to 3eV/e​=3V
So, the answer is −3V i.e. option B is correct.

The number of photoelectrons emitted per second from a photosensitive plate is directly proportional to the intensity of the incident radiation. ... For the same frequency of light and increased intensity, the saturation current is found to increase, but the cut-off potential is found to remain constant.
The number of electrons also changes because of the probability that each photon results in an emitted electron are a function of photon energy. If the intensity of the incident radiation of a given frequency is increased, there is no effect on the kinetic energy of each photo electron.

The answer is option A. 
First, the gradient of the graph cannot change (always = h/e ), so answers are A or D.
If Y has a greater work function than X, then the graph for Y should have a more negative y-intercept. therefore figure in option A depicts the concept.

Lenses of the same diameter collect equal amounts of light.
Intensity is the measure of amount of light collected, hence the intensity remains the same.
Intensity is measured by the photoelectric current. So the photoelectric current would also remain the same.

Charge present on both particles are same
∵λ= h​/mqV so q₁​=q₂​
λ_e​​/λ_p​ = √(m_p​v​​/m_e​v) =>1
∵m_p​>>m_e​
∴λ_e​>λ_p​.
Option C is the correct answer.

Momentum of photo, p=E/c​=hν/c ​where E is the energy of a photon and c is the velocity of light. 
∴ p= hc/cλ ​    [∵ν=λc​]
p=h/λ​=h/(0.01×10−10)​=10¹²h

Angular momentum,
L=nh/2π​=3h​/2π (given)
also n=3
∴  E=(−1.36​/ n²)eV
E=−13.6/(3)^2​ eV
E=−13.6/9 ​eV
E=−1.57eV

ΔE = k[(1/n₂²)-1/n₁²] = hc / λ

from n = 4 → n = 2
ΔE = maximum energy
wavelength λ will be shortest for C
If transition occurs from n = 4 → n = 3
then ΔE→ minimum
λ will be longest for D.

C→ Shortest
D→ longest

Energy of e− in n^th shell =−13.6 (τ²​ /n²)
For H−atom with n=3→E=−13.6×(1/9)​
⇒E=1/9​ (Energy of e− in first shell)
⇒E=(1/9)​×2.18×10⁻¹⁸)
⇒E=2.42×10⁻¹⁹J(Ionisation Energy= Energy of e⁻ in first shell)

We know that r_n​∝n²
Given: r_n+1​−r_n​=r_n−1​
∴(n+1)²−n²=(n−1)²
n=4

According to Bohr's atomic theory,
radius of nth orbit, r ∝ n²/Z
velocity of electron in nth orbit, v ∝ Z/n
then acceleration of electron in nth orbit, a = v²/r
= (Z/n)²/(n²/Z)
= Z3/n4
so acceleration, a ∝ 1/n⁴
now for m shell, n = 3
for l shell, n = 2
now initial acceleration/final acceleration = a₁/a₂
= (n²/n1)4 = (2/3)⁴ = 16/81
Therefore the ratio of initial to final acceleration is 16:81.
 

frequency∝ z²​/n³
⇒f=kz²/n^3​
∴f1​=(1/27) f₂​
⇒​kz₁²/n₁³ ​​=(1/27) (kz₂²/n₂³​)   ​​(z₁​=z₂​=1)
⇒k/n₁³​​=(1/27) (k/n₂³)
⇒n₁​=3n_2​
∴n₁​=3 and n_2​=1

Radius of nth orbital
r_n​= ϵ₀​n₂h₂​/ πmZe₂
Wherein 
=r_n​∝ n²​ /Z
= ϵ_0​h²/πme² ​=0.529
r=(n^2​/Z)a₀​
For Z=1 r=n²a₀​

Ionization energy-
E_ion=E_∞-E_n
=13.6 (Z²/n²)
- wherein
Energy required to move an electron from ground state to 
n=∞
Energy required to remove electron from 2nd bohr orbit is 
△E=E₀ . [1/{2²-(1/∞)}]=E₀/4=(13.6/4) eV
△E=3.4eV

Given,
n_1​=1
n_2​=5
m=1.6×10⁻²⁷kg
Momentum, P=mv=h​/λ
recoil speed, v= h​/mλ. . . . . . . . .(1)
we know that,
1/λ​=R[(1/n₁²)​​−(1/n₂²​​)]
1/λ ​=1.097×107[(1/1²​)−(1/5^2​)]
λ=9.48×10⁻⁸m
From equation (1),
recoil speed, v=6.67×10^−34​/1.6×10⁻²⁷×9.48×10⁻⁸ =4.2m/s
The correct option is C.
 

Wavelength is calculated as 
1/λ=Rz²(1/n_f²​​=1/n_i²​​)
∴λ∝1/z²
For hydrogen atom, z=1
For Deuterium atom, z=2
For singly ionized helium, z=2
For doubly ionized lithium, z=3
Thus, z is maximum for lithium and so λ is minimum.
Hence the correct answer is option D.

Time period, T∝(R)3/2
∴ T1/T2​ ​​= (​R1/R2 ​)3/2= (1/4​)3/2=1/8
Thus, the ratio of time period is 1:8