Dual Nature of Matter and Radiation Test - 8

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 ×10⁻¹⁹ C ​=10¹² electrons are generated per second
Now we notice that one photon has energy E, E=hc/λ=​=(6.62 ×10⁻³⁴ Js ×3 ×10⁸ ms^-1 )/(5000 ×10 −10m) ​=3.972 ×10⁻¹⁹ Joule
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 ​} =h ν−ϕ
where, νis the frequency of incident light and ϕis photoelectric work function of metal.
eV_{0 ​} =h ν−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 ′​=2h ν−ϕ 
eV₀ ′​=2[h ν−(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/e λ)-(w/e)
Arithmetic progression =>V₂ =(V₁ +V₂ )/2
Now,
(hc/e λ₂ )-w/e=1/2[(hc/e λ₁ )-(w/e) +(hc/e λ₃ ) -(w/e)]
=>1/ λ₂ =1/2[(1/ λ₁ )+(1/λ₃ )]
=>2/ λ₂ =1/ λ₁ + 1/λ₃
Hence the correct answer is harmonic Progression.