Question

Radiations of wavelength $5000\mathring{A}$ fall on a metal plane whose work function is $1.9eV$. Calculate the stopping potential

• A. $0.97volt$
• B. $0.17volt$
• C. $0.37volt$
• D. $0.57volt$

Answer 1

The correct option is D $0.57volt$

We know, $E=kE+h{V}_0⟶\left(1\right)$

$E=hv=\left(h_c\right)/\lambda$

$E=(6.63\times {10}^{-34}\times 3\times {10}^{-8})/(5000\times {10}^{-10})$

$=3.978\times {10}^{-19}J$

$1eV=1.6\times {10}^{-19}J$

$1J=1/(1.6\times {10}^{-19})eV$

$\therefore$ $3.978\times {10}^{-19}J=(1\times 3.978\times {10}^{-19})/(1.6\times {10}^{-19})$

$=2.486eV$

Thus, $KE=E-h{v}_0$

$KE=2.486-1.9=0.586eV$

Kinetic energy of photon $=0.586eV$