Question

The work function of aluminum is 4.2eV. Light of wavelength $2000\mathring{A}$ is incident on it. The threshold wavelength will be

• A. $5868\mathring{A}$
• B. $2946\mathring{A}$
• C. $3856\mathring{A}$
• D. $5000\mathring{A}$

Answer 1

The correct option is B $2946\mathring{A}$

Threshold frequency is defined as the minimum frequency of incident light which can cause photo electric emission i.e. this frequency is just able to eject electrons without giving them additional energy. The energy corresponding to this frequency is the photoelectric work function of the metal.
${\varphi }_0=h{\nu }_0$

${\varphi }_0=\frac{hc}{{\lambda }_0}$

${\lambda }_0=\frac{hc}{{\varphi }_0}$

${\lambda }_0=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{4.2\times 1.6\times {10}^{-19}}$

${\lambda }_0=\frac{19.8\times {10}^{-26}}{6.72\times {10}^{-19}}$

${\lambda }_0=2.946\times {10}^{-7}$
${\lambda }_0=2946\times {10}^{-10}m$
${\lambda }_0=2946A^{\circ }$

So, the answer is option (B).