Question

A photon of wavelength $5000A^{\circ }$ strikes a metal surface having work function of $2.20eV$. The kinetic energy of the emitted photo electron is?

• A. $4.4\times {10}^{-20}J$
• B. $0.425eV$
• C. $2.20eV$
• D. No electron will be emitted

Answer 1

Answer: (A)

$4.4\times {10}^{-20}J$

(1)We know $\lambda =5\times {10}^{–7}$ m(given)

$C=3\times {10}^8$

From the equation $E=hv$ or$hc/\lambda$

Where, h = Planck’s constant $=6.626\times {10}^{–34}$ Js

c = velocity of light in vacuum $=3\times {10}^8$ m/s

$\lambda =$ wavelength of photon $=5\times {10}^{–7}$ m

Substituting the values in the given expression of E:

$E=(6.26\times {10}^{-34})(3\times {10}^8)/5\times {10}^{-7}=3.9695\times {10}^{-19}J$

Hence, the energy of the photon is $3.97\times {10}^{–19}$ J.

(ii) The kinetic energy of emission Ek is given by

$=hv-h{v}_0$

$(E-W)eV$

$\frac{3.9695\times {10}^{-19}}{1.6020\times {10}^{-19}}eV-2.20eV$

$=(2.481–2.20)eV$

$=0.281eV$

$1eV=1.6\times {10}^{-}19J$

so $0.281eV=0.281\times 1.6\times {10}^{-19}=4.48\times {10}^{-20}$