Question

The work function of a metal is 1.5 eV. Light of wavelength $6600\mathring{A}$ is made incident on it. The maximum kinetic energy of emitted photoelectrons will be

• A. $1.6\times {10}^{-19}Joule$
• B. $0.6\times {10}^{-19}Joule$
• C. $1.6\times {10}^{-13}Joule$
• D. $1.6\times {10}^{19}Joule$

Answer 1

The correct option is B $0.6\times {10}^{-19}Joule$

The maximum kinetic energy for the photoelectrons is
$E_{max}=h\nu -\varphi$
where,$\nu$ is the frequency of incident light and $\varphi$ is photoelectric work function of metal.

$E_{max}=h\frac{c}{\lambda }-\varphi$ .....................(since, $\nu =\frac{c}{\lambda }$)

$E_{max}=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{6600\times {10}^{-10}}-1.5\times 1.6\times {10}^{-19}$

$E_{max}=3\times {10}^{-19}-2.4\times {10}^{-19}$

$E_{max}=0.6\times {10}^{-19}J$

So, the answer is option (B).