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  • What is the work function of a metal if it emits electrons with a kinetic energy of 2.5 eV when exposed to light with a wavelength of 500 nm? (Given: <img alt="h= 6.626 \times 10^{-34}J.s

What is the work function of a metal if it emits electrons with a kinetic energy of 2.5 eV when exposed to light with a wavelength of 500 nm?

(Given: h= 6.626 \times 10^{-34}J.s,c= 3.0\times10^{8}m/s,1eV= 1.6\times 10^{-19}J)

Option: 1

3.94 eV


Option: 2

4.47 eV


Option: 3

2.13 eV


Option: 4

1.48 eV


Answers (1)

best_answer

The energy of the incident light can be calculated using the equation

E= \frac{hc}{\lambda }.
E= \frac{\left ( 6.626 \times 10^{-34}J.s \right )\times\left ( 3.0\times 10^{8}m/s \right )}{\left ( 500\times10^{-9}m \right )}
E = 3.98 eV
The work function of the metal can be calculated using the equation

\phi= E-KE,
where KE is the kinetic energy of the emitted electrons.
\phi= 3.98eV-2.5eV= 1.48eV

Posted by

Ritika Kankaria

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