Question

The threshold wavelength for tungsten is $272nm$. A light of frequency $1.45\times {10}^{15}Hz$ is incident. Find the stopping potential.

• A. $1.37V$
• B. $1.47V$
• C. $1.57V$
• D. $1.51V$

Answer 1

Answer: (B)

$1.47V$

Since the threshold wavelength is 272nm then the work function is -

$\varphi =\frac{1240}{272}=4.56eV;$

Energy of the incident photon is-

$h\nu =\frac{6.625\times {10}^{-34}\times 1.45\times {10}^{15}}{1.6\times {10}^{19}}$

$=6.03eV$
So the stopping potential is-
$e{V}_s=h\nu -\varphi$ or $V_s=6.03-4.56=1.47V$
So, the answer is option (B).