Question

A photon with initial frequency ${10}^{11}$ Hz scatters off an electron at rest. Its final frequency is 0.9 X ${10}^{11}$ Hz. The speed of scattered electron is close to $(h=6.63\times {10}^{-34}$ J s, $m_e=9.1\times {10}^{-31}$ kg)

• A. 4 X 10³ ms^-1
• B. 3 X 10² ms^-1
• C. 2 X 10⁶ ms^-1
• D. 30 ms^-1

Answer 1

The correct option is A 4 X 10³ ms^-1

$hv=h{v}_0+\frac{1}{2}m{v}^2$
$\frac{1}{2}m{v}^2=h(v-v_0)$
$v^2=\frac{2h(v-v_0)}{m}$
$=\frac{2\times 6.63\times {10}^{-34}({10}^{11}-0.9\times {10}^{11})}{9.1\times {10}^{-31}}$
$v=\sqrt{14571428}=3.8\times {10}^3\approx 4\times {10}^{3}m{s}^{-1}$