Question

Light of wavelength 200 nm is incident on a cadmium surface. A stopping voltage of 2.15 ev is required to reduce the photocurrent to zero. What is the work function of cadmium?

a. 2.15 ev

b. 6.22 ev

c. 8.37 ev

d. 4.07 ev 2 points

Answer 1

eV=hc/Lamda -W

All Variables are given

V=2.15eV

Lamda =200 nm

e=1.6×10^-19

Just equate for W .

Answer 2

The work function of cadmium is 2.15 eV

Explanation:

Given that,

Wavelength, $\lambda=200\ nm=2\times 10^{-7}\ m$

Stopping voltage, $E=2.15\ ev=2.15\times 1.6\times 10^{-19}\ J$

Using the equation of photoelectric equation as :

$\phi=\dfrac{hc}{\lambda}-E$

$\phi=\frac{6.63\cdot10^{-34}\cdot3\cdot10^{8}}{2\cdot10^{-7}}-2.15$

$\phi=2.15\ eV$

So, the work function of cadmium is 2.15 eV. Hence, this is the required solution.

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Topic : Photoelectric effect

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