Dual Nature of Radiation and Matter Test 5

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=>K₁2/2

• 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

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

Correct Answer :- d

Explanation : Statement I is incorrect

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

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

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