Question

The wavelength of the photo-electric threshold of a metal is $230nm$. Determine the energy of the electrons ejected from the surface by $UV$ light of $\lambda$ $=180nm$.
(Given $h=6.626\times {10}^{-27}erg-sec$)

• A. $2.36\times {10}^{-19}J$
• B. $2.44\times {10}^{-22}kJ$
• C. $2.188\times {10}^{-17}J$
• D. $3.975\times {10}^{-18}J$

Answer 1

The correct option is A $2.36\times {10}^{-19}J$

The energy of the UV radiation is $E=\frac{hc}{\lambda }=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{180\times {10}^{-9}}=1.1\times {10}^{-18}$ J.
The workfunction of the metal is $E_0=\frac{hc}{{\lambda }_0}=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{230\times {10}^{-9}}=8.64\times {10}^{-19}$ J
The energy of the ejected electron is $E-E_0=1.1\times {10}^{-18}-8.64\times {10}^{-19}=2.36\times {10}^{-19}$ J