Question

The minimum energy required by an electron to overcome the attractive forces of the nucleus is $0.4\times {10}^{-19}\text{J}$. What is the maximum kinetic energy of the ejected electron if the material is radiated with a light of wavelength $3313\text{nm}$?

• A. $0.2\times {10}^{-20}\text{J}$
• B. $2\times {10}^{-20}\text{J}$
• C. $1\times {10}^{-20}\text{J}$
• D. $10\times {10}^{-20}\text{J}$

Answer 1

The correct option is B $2\times {10}^{-20}\text{J}$

Step 1: Given wavelength of the incident radiation $\left(\lambda \right)=3313\text{nm}$
Also, the minimum energy required by the electron to overcome the attractive forces ( $E_0$) $=0.4\times {10}^{-19}\text{J}$

Step 2: Finding the energy of the incident radiation $\left(E\right)$
We know that, $E=\frac{hc}{\lambda }$
Where, $h=$ Planck's constant $=6.626\times {10}^{-34}\text{Js}$
$c=$ speed of light $=3\times {10}^8\text{m/s}$
$\lambda =$ Wavelength of the incident radiation
$=3313\text{nm}$
$=3313\times {10}^{-9}\text{m}$

$\therefore E=\frac{6.626\times {10}^{-34}\text{J.s}\times 3\times {10}^8\text{m/s}}{3313\times {10}^{-9}\text{m}}$
$E=(\frac{6626\times {10}^{-3}}{3313\times {10}^{-9}}\times {10}^{-34}\times 3\times {10}^8)\text{J}$
$=6\times {10}^{-20}\text{J}$
$E=0.6\times {10}^{-19}\text{J}$

Step 3: According to Einstein's photoelectric equation,
$E=E_0+K.E_{\text{max}}$
Where, $E$ is the energy of the incident light
$E_0$ is the work function.
$K{E}_{max}$ is the maximum kinetic energy of the ejected electron.
Substituting the values of $E$ and $E_0$,
$0.6\times {10}^{-19}\text{J}=0.4\times {10}^{-19}\text{J}+K.E_{\text{max}}$
$\therefore K.E_{\text{max}}=(0.6-0.4){10}^{-19}\text{J}$
$=0.2\times {10}^{-19}\text{J}$
K.E${}_{\text{max}}=2\times {10}^{-20}\text{J}$