Test: Dual Nature of Radiation of Matter- 2 (March 8) - NEET 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%

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.

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

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 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.