A metal surface is illuminated by a radiation of wavelength 4500 Å. The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90° with the magnetic field. If it starts revolving in a circular path of radius 2 mm, the work function of the metal is approximately :

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2.78 eV

2.23 eV

1.36 eV

1.69 eV

answer is A.

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Detailed Solution

$λ = 4500 Å$
B = 2mT, R = 2mm
$\begin{array}{l} R = \frac{\sqrt{2 Km}}{qB} \\ \frac{(qBR)^{2}}{2 m} = K \end{array}$

$\begin{array}{l} \frac{{(1 .6 \times 10^{- 19} \times 2 \times 10^{- 3} \times 2 \times 10^{- 3})}^{2}}{2 \times 9 .1 \times 10^{- 31}} = K \\ \frac{(6 .4)^{2}}{2 \times 9 .1} \times \frac{10^{- 50}}{10^{- 31}} = K \end{array}$
K = 2.25 × 10^–19 J
$\begin{array}{l} = \frac{2 .25 \times 10^{- 19}}{1 .6 \times 10^{- 19}} eV = 1 .40 eV \\ E = \frac{12400}{4500} = 2 .76 eV \end{array}$
$ϕ$ = E – K = (2.76 – 1.40) eV = 1.36 eV