Question

The threshold wavelength for ejection of an electron from metal is 310nm. The kinetic energy of emitted photoelectron is $36\times {10}^{-20}J$. Then the frequency of incident light is:

• A. $1.5\times {10}^{15}Hz$
• B. $2.5\times {10}^{18}Hz$
• C. $1\times {10}^{22}Hz$
• D. $2.5\times {10}^{12}Hz$

Answer 1

The correct option is A $1.5\times {10}^{15}Hz$

Use,
$K.E.=h\upsilon -\varphi$
We have been given threshold wavelength which can be used to calculate work function by,
$\varphi =hc/{\lambda }_{threshold}$
$\therefore h\upsilon =K.E.+\varphi =36\times {10}^{-20}+63.9\times {10}^{-20}J=99.9\times {10}^{-20}J$
$\therefore \upsilon =99.9\times {10}^{-20}/6.6\times {10}^{-34}=1.5\times {10}^{15}Hz$.