Question

A certain metal when irradiated by a light of frequency $\nu =3.2\times {10}^{16}\text{Hz}$ emits photoelectrons with twice the kinetic energy of photoelectrons that are emitted when the same metal is irradiated by a light of frequency $\nu =2.0\times {10}^{16}\text{Hz}$. Calculate the work function of the metal.

• A. $6.0\times {10}^{15}$ Hz
• B. $8.0\times {10}^{17}$ Hz
• C. $8.0\times {10}^{15}$ Hz
• D. $6.0\times {10}^{17}$ Hz

Answer 1

The correct option is C $8.0\times {10}^{15}$ Hz

Applying photoelectric equation,
$KE=h\nu -h{\nu }_0$
Where ${\nu }_0$ = threshold frequency
$\nu$ = frequency of incident radiation
h = Planck's constant
$K.E.$ = Planck's constant
$\Rightarrow (\nu -{\nu }_0)=\frac{KE}{h}$

Given: $K{E}_2=2K{E}_1$

$({\nu }_2-{\nu }_0)=\frac{K{E}_2}{h}...\left(i\right)$

and $({\nu }_1-{\nu }_0)=\frac{K{E}_1}{h}...\left(ii\right)$

Dividing equation (i) by (ii),
$\frac{{\nu }_2-{\nu }_0}{{\nu }_1-{\nu }_0}=\frac{K{E}_2}{K{E}_1}=\frac{2K{E}_1}{K{E}_1}=2$

$\Rightarrow {\nu }_2-{\nu }_0=2{\nu }_1-2{\nu }_0$
$\Rightarrow {\nu }_0=2{\nu }_1-{\nu }_2=2(2\times {10}^{16})-(3.2\times {10}^{16})$
$=8\times {10}^{15}\text{Hz}$