Test: Dual Nature of Radiation & Matter - 3 - JEE MCQ

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.

We know that,
[(hc/ λ) -W_o]=K₁ =>l₁=hc/(K₁ +W_o)
[(hc/l₂ )-Wo]=K₂=>l₂= hc/(K2+ W_o)
Given, l₁=2l₂
=> hc/K₂+ W_o =2hc/(K₂+ W_o)
Or,K₁+Wo= (K₂+ W_o)/2
Or,K₁=(K₂/2)-(Wo/₂)
Since, W_o >1=>K₁<K₂/2

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)
   

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 

Given, incident wavelengths travel in x-direction which are in the U.V region, which means they can excite an electron from n=1 to higher levels. Hence, some of the incident wavelengths will be absent in A.
Photons emerging due to transitions of electrons to their ground state contains visible region wavelengths if transition is done into n=2. If transition is done into n>2, photons contain I.R wavelengths.
Therefore, A,C,D is the correct choice.
 

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.

Correct Answer :- a

Explanation : When the electron changes levels, it decreases energy and the atom emits photons. The photon is emitted with the electron moving from a higher energy level to a lower energy level. The energy of the photon is the exact energy that is lost by the electron moving to its lower energy level.

Correct Answer :- d

Explanation : a) Metals emit electrons when light shines upon them. The phenomenon is named as photoelectric effect and the electrons emitted in such manner are called photoelectrons.

The kinetic energy of photoelectrons vary simply because of the fact that (after photoelectric effect has taken place) all the photoelectrons are not emitted instantly. Most of the photoelectrons first collide (several times!) with the other electrons (which are roaming around with random velocities; recall that metals are generally electrically conductive i.e. having abundant free electrons) within the metal. These photoelectrons lose some fraction of their kinetic energy in such collisions before they are emitted out of the metal.

That's why it is observed that the photoelectrons do not have the same kinetic energy.

b) In the photoelectric effect, each photon donates all of its energy hf to an electron in the metal. If this process occurs at the metal surface, the electron is released into the vacuum with a kinetic energy given by:

Kmax = hf - phi

de Broglie Wavelength for Electrons
In the case of electrons going in circles around the nuclei in atoms, the de Broglie waves exist as a closed-loop, such that they can exist only as standing waves, and fit evenly around the loop. Because of this requirement, the electrons in atoms circle the nucleus in particular configurations, or states, which are called stationary orbits.
de Broglie Wavelength Formula and Derivation
de Broglie reasoned that matter also can show wave-particle duality, just like light, since light can behave both as a wave (it can be diffracted and it has a wavelength) and as a particle (it contains packets of energy hν). And also reasoned that matter would follow the same equation for wavelength as light namely,
λ = h / p
Where p is the linear momentum, as shown by Einstein.
Derivation
de Broglie derived the above relationship as follows:
1) E = hν for a photon and λν = c for an electromagnetic wave.
2) E = mc², means λ = h/mc, which is equivalent to λ = h/p.
Note: m is the relativistic mass, and not the rest mass; since the rest mass of a photon is zero.
Now, if a particle is moving with a velocity v, the momentum p = mv and hence λ = h / mv
Therefore, the de Broglie wavelength formula is expressed as;
λ = h / mv

Correct Answer :- d

Explanation : Statement I is incorrect

The correct statement is : Two photons having Equal linear momenta have equal wavelengths.

I=nhv/ΔtA=[N.h(C/ λ)]/(1mm/c)x(1mm2)
N=200x10^-9x2640x10^-10/(3x10⁸)²x6.63x10^-34
   =885 Photons/mm³

We know that,
∣P.E∣/2​=∣T.E∣=K.E
So, ∣P.E∣/2​=3.4=K.E
So, kinetic energy    K.E=3.4 eV

Intensity of photons ↑, no. of electron ↑
 
The cut off voltage does not depend on distance between cell and source.
 But intensity is inversely proportional to square of it's distance. 
∴ at d=0.6m,
Cut off voltage =0.6V
 
but 
I_s​=18×(0.2/0.6 ​)²
=2mA

Intensity of photons ↑, no. of electron ↑
The cut off voltage does not depend on distance between cell and source.
 But intensity is inversely proportional to square of it's distance. 
∴ at d=0.6m,
Cut off voltage =0.6V
but 
I_s​=18×(0.2/0.6​)2
=2mA
So, the answers are option 2.0 mA
 

By Einstein's Photoelectric equation
eV=hν−ϕ
where V=1V= cut off voltage
⟹ν= (eV+ϕ)/ h ​=4eV/h​=4×1.6×10^−19​/6.63×10⁻³⁴
⟹ν=9.6×10¹⁴Hz

N / Δt = no. of photon incident per second.
∴ 663x10^-3=(N/Δt)x(hc/λ)
∴N/ Δt=663x10^-3/(hc/λ)= 663x10^-3/(1242nmeV/540)
(n/Δt)/(N/Δt)=1/(5x10⁹)
n/Δt=[1/(5x10⁹)]x (N/Δt)
       =[1/(5x10⁹)]x ((663x10^-3x540)/1242nmeV)
I=ne/Δt=5.76x10¹¹A