A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are $u_{1} a n d u_{2},$ respectively. If the ratio $u_{1} : u_{2} = 2 : 1$ and $h c = 1240 eV-nm,$ the work function of the metal is nearly,

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$3.2 e V$

$2.8 e V$

$2.5 e V$

$3.7 e V$

answer is A.

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Detailed Solution

Energy corresponding to $248 n m$ wavelength
$E = \frac{1240}{λ} e V = \frac{1240}{248} e V = 5 e V$
Energy corresponding to $310 n m$ wavelength $= \frac{1240}{310} e V = 4 e V$
$K E = E - W o r k f u n c t i o n ∴ \frac{K E_{1}}{K E_{2}} = \frac{u_{1}^{2}}{u_{2}^{2}} = \frac{4}{1} = \frac{5 e V - W}{4 e V - W}$

solving this equation, we get $W = 3.67 e V ≅ 3.7 e V$

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