Question

Energy required to stop the ejection of electron from $Cu$ plate is $0.24eV$. If radiation of $\lambda =253.7nm$ strikes the plate, the work function is:

• A. $4.65eV$
• B. $4.89eV$
• C. $4.24eV$
• D. $3.0eV$

Answer 1

The correct option is B $4.89eV$

Energy of photon =work function = $12m{v}^2$

Energy of photon =work function = $e{v}_0$------(1)

Where e is electronic charge and v0 is stopping potential and $e{v}_0$ is equal to energy required to stop the ejection of electron.

$E_{photon}=hc/\lambda$

$=\frac{6.625\times {10}^{-}34\times 3.0\times {10}^8}{253.7\times {10}^{-9}}$

$=7.834\times {10}^{-19}J$

$=\frac{7.834\times {10}^{-19}}{1.602\times {10}^{-19}}eV=4.89eV$