Question

The stopping potential for the photo electrons emitted from a metal surface of work function $1.7eV$ is $10.4V$. 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

Answer 1

The correct option is A n = 3 to 1

As we know that the stopping potential of the photoelectron is equal to the maximum kinetic energy of the photoelectron,

$K{E}_{max}=10.4V$

Now, in photoelectric effect,

Energy of incident radiation $\left(E_{in}\right)$ = work function + $K.E_{max}$

$\Rightarrow E_{in}=1.7+10.4$

$\Rightarrow E_{in}=12.1eV$

Now, for 0 hydrogen atom,

Energy of first energy level, $E_1=-13.6eV$

Energy of second energy level , $E_2=-3.4eV$

Energy of third energy level, $E_3=-1.5eV$

Hence, a transition from third to first energy level will result in emission of radiation of energy = $E_3-E_1=12.1eV$ which is same as the energy of incident radiation of above photoelectric effect.

Thus, correct answer is $n=3$ to $1$