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The stopping potential for the photo electrons emitted from a metal surface of work function 1.7eV is 10.4 V. Identify the energy levels corresponding to the transitions in hydrogen atom which will result in emission of wavelength equal to that of incident radiation for the above photoelectric effect
  • a)
    n = 3 to 1
  • b)
    n = 3 to 2
  • c)
    n = 2 to 1
  • d)
    n = 4 to 1
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The stopping potential for the photo electrons emitted from a metal su...
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 (Ein​) = work function + K.Emax​
⇒ Ein​=1.7+10.4
⇒ Ein​=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
 
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Explanation:

The given stopping potential is the minimum potential required to stop the photoelectrons emitted from a metal surface. The stopping potential is given by the equation:

Stopping potential = (hc/λ) - work function

where h is Planck's constant, c is the speed of light, λ is the wavelength of incident radiation and work function is the minimum energy required to remove an electron from the metal surface.

Using the given values, we can calculate the wavelength of the incident radiation as:

λ = hc/(Stopping potential + work function)
= 6.63 x 10^-34 J s x 3 x 10^8 m/s / (10.4 V + 1.7 eV)
= 4.6 x 10^-7 m

This wavelength corresponds to the transition of an electron from a higher energy level to a lower energy level in a hydrogen atom, emitting a photon of the same wavelength.

The energy levels of a hydrogen atom are given by the equation:

En = (-13.6 eV) / n^2

where En is the energy of the nth level and n is the principal quantum number.

To find the transition corresponding to the given wavelength, we can use the equation:

ΔE = hc/λ = Ei - Ef

where ΔE is the energy difference between the initial (Ei) and final (Ef) energy levels, h is Planck's constant and c is the speed of light.

Substituting the given values, we get:

ΔE = 6.63 x 10^-34 J s x 3 x 10^8 m/s / (4.6 x 10^-7 m)
= 4.3 x 10^-19 J

This energy difference corresponds to the transition from the third energy level (n=3) to the first energy level (n=1) in a hydrogen atom, which emits a photon of wavelength 4.6 x 10^-7 m.

Therefore, the correct option is A, which corresponds to the transition from n=3 to n=1 in a hydrogen atom.
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The stopping potential for the photo electrons emitted from a metal surface of work function 1.7eV is 10.4 V. Identify the energy levels corresponding to the transitions in hydrogen atom which will result in emission of wavelength equal to that of incident radiation for the above photoelectric effecta)n = 3 to 1b)n = 3 to 2c)n = 2 to 1d)n = 4 to 1Correct answer is option 'A'. Can you explain this answer?
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