Question

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

Answer 1

The correct option is A n = 3 to 1

As we know that the maximum kinetic energy of the photoelectron is equal to e times the stopping potential of the photoelectron
$K{E}_{max}=10.4\text{eV}$
Now, in photoelectric effect, Energy of incident radiation
$\left(E_{in}\right)$ = work function + $K{E}_{max}$
$\Rightarrow E_{in}=1.7+10.4$
$\Rightarrow E_{in}–12.1\text{eV}$
Now, for hydrogen atom,
Energy level, $E_n=\frac{-13.6}{n^2}$
Energy of first energy level, $E_1=-13.6\text{eV}$
Energy of second energy level, $E_2=-3.4\text{eV}$
Energy of third energy level, $E_3=-1.5\text{eV}$
Hence, a transition from third to first energy level will result in emission of radiation of energy $=E_3–E_1=12.1\text{eV}$ which is same as the energy of incident radiation of above photoelectric effect.
So, the correct answer is n = 3 to n = 1