Question

Threshold wavelength of a metal is $4000\stackrel{o}{A}$. If light of wavelength $3000\stackrel{o}{A}$ irradiates the surface, the maximum kinetic energy of photoelectron is?

• A. $1.7$eV
• B. $1.6$eV
• C. $1.5$eV
• D. $1.0$eV

Answer 1

Answer: (D)

$1.0$eV

If the maximu kinetic energy of photoelectron emitted from the surface of a metal is $E_k$ and $W$ is work function of the metal, the from Einstein photoelectric equation we have
$E_k=hv-W$
where,hv is the energy of the photon absorbed the electron in the metal.
Also $v=\frac{c}{\lambda }$
$=\frac{velocity}{wavelength}$
$=E_k=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}=hc\left(\frac{1}{\lambda }-\frac{1}{{\lambda }_0}\right)$
Given, $\lambda =3\times {10}^{-7}$m,
${\lambda }_0=4\times {10}^{-7}$m.
$\therefore E_k=.6\times {10}^{-34}\times 3\times {10}^8\times \left(\frac{1}{3\times {10}^{-7}}-\frac{1}{4\times {10}^{-7}}\right)$
$=\frac{19.8\times 10-19}{12\times 1.6\times {10}^{-19}}$eV
$=1.03\approx 1$eV.