Question

A beam of $ 450\, nm $ light is incident on a metal having work function $ 2 \,eV $ and placed in a magnetic field $ B $ . If the most energetic electrons emitted are bent into circular are of radius $ 0.2\, m $ , find $ B $ .

• A. $ 2.36 \times 10^{-4}\, T $
• B. $ 1.46 \times 10^{-5}\, T $
• C. $ 6.9 \times 10^{-5}\, T $
• D. $ 9.2 \times 10^{-6}\, T $

Answer 1

Answer: (D)

$\frac{1}{2} mv^2 + \phi = E$
$= h \frac{c}{\lambda}$
$\frac{1}{2}mv^2 + 2 \times 1.6 \times 10^{-19} = 6.6 \times 10^{-34}$
$\times \frac{3\times 10^8}{450 \times 10^{-9}}$
$ mv^2 = \frac{6.6 \times 10^{-34} \times 3 \times 10^8 \times 2}{450 \times 10^{-9}}$
$ - 4 \times 1.6 \times 10^{-19}$
$\frac{1}{2} \times 1.9 \times 10^{-31} v^2 = 8.8 \times 10^{-19}$
$ - 4 \times 1.6 \times 10^{-19}$
$v^2 = \frac{4.8 \times 10^{-19}}{1.9\times 10^{-31}}$
$ v = 1.58 \times 10^6$