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Q.

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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a

$3.2 e V$

b

$2.8 e V$

c

$2.5 e V$

d

$3.7 e V$

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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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 u1 and u2, respectively. If the ratio u1 :u2=2:1 and hc=1240 eV-nm, the work function of the metal is nearly,