Question

The work function of a surface of a photosensitive material is 6.2 eV. The wavelength of the incident radiation for which the stopping potential is 5V lies in the :
A. infrared region
B. X-ray region
C. Ultraviolet region
D. Visible region

Answer 1

We can find the wavelength of the incident radiation by making use of the photoelectric equation according to which energy of the incident radiation is equal to the sum of the work function and the stopping potential in eV. Then by analysing the value of the wavelength, we can find out the region of electromagnetic spectrum to which that wavelength belongs.
Formula used:
The formula for photoelectric effect is given in terms of the work function and stopping potential by the following equation:
$\dfrac{{hc}}{\lambda } = W + eV$

Complete step by step answer:
We are given the value of the work function of a surface of a photosensitive material which is
$W = 6.2eV$
The stopping potential for a certain wavelength is given as
$V = 5V$
The formula for photoelectric effect is given in terms of the work function and stopping potential by the following equation:
$h\nu = W + eV$
where $h\nu$ is the energy of incident radiation; h is called the Planck’s constant which has value $h = 6.62 \times {10^{ - 34}}Js$ and $\nu$ is the frequency of radiation which is related to wavelength $\lambda$ as follows:
$\nu = \dfrac{c}{\lambda }$
Here c is the velocity of light given as $c = 3 \times {10^8}m/s$. Also $1eV = 1.6 \times {10^{ - 19}}J$
Let us find out the wavelength of the radiation from the given values of work function and the stopping potential. It can be done in the following way.
$\dfrac{{hc}}{\lambda } = W + eV = 6.2eV + 5eV = 11.2eV \\\ \Rightarrow \lambda = \dfrac{{hc}}{{11.2eV}} \\\ = \dfrac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{11.2 \times 1.6 \times {{10}^{ - 19}}}} \\\ = 1.108 \times {10^{ - 7}}m = 110.8nm \\\$
This wavelength lies in the ultraviolet region.

Therefore, the correct answer is option C.

Note:
It is necessary to remember the ranges of wavelengths for different radiation in the electromagnetic spectrum in order to answer this question. The approximate values of wavelengths for different radiations are given as follows:
Gamma rays: ${10^{ - 6}}$ nm
X-rays: 1 nm to 10 nm
Ultraviolet: 100 nm
Visible: 400 nm to 800 nm
Infrared: 10000 nm
Microwaves: 1 cm
Radio waves: 10 cm