Question

The work function of aluminum is 4.2eV. Light of wavelength $2000\mathring{A}$ is incident on it. The stopping potential for fastest electrons will be

• A. Zero
• B. $2.025$ volt
• C. $4.2$ volt
• D. $6.19$ volt

Answer 1

Answer: (B)

$2.025$ volt

Energy of incident photons $E=\frac{hc}{\lambda }=\frac{6.64\times {10}^{-34}\times 3\times {10}^8}{2000\times {10}^{-10}}=9.96\times {10}^{-19}J$

The energy in electron volts is: $\frac{E}{1.618\times {10}^{19}}=6.2250eV.$

Hence the energy with which the electrons are ejected is :
$6.225-4.2=2.025eV$
That is the stopping potential needed to stop the fastest electron.