Question

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

• A. Zero
• B. 1.99 volt
• C. 3.2 volt
• D. 6.19 volt

Answer 1

Answer: (B)

1.99 volt

The work function of aluminium is
$\varphi =4.2eV=4.2\times 1.6\times {10}^{-19}$
$\varphi =6.72\times {10}^{-19}J$
The energy of incident photon is

$h\nu =\frac{hc}{\lambda }$

$h\nu =\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{2000\times {10}^{-10}}$

$h\nu =9.9\times {10}^{-19}J$
The maximum energy of photoelectrons is
$E_{max}=h\nu -\varphi$
$E_{max}=9.9\times {10}^{-19}-6.72\times {10}^{-19}$
$E_{max}=3.18\times {10}^{-19}J$

$E_{max}=\frac{3.18\times {10}^{-19}}{1.6\times {10}^{-19}}$

$E_{max}=1.99eV$

So, the answer is option (B).