Ncert Exercises & Solutions

2.52 Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant.

λ (nm)	500	450	400
v × 10–5 (cm s–1)	2.55	4.35	5.35

(a) Assuming the threshold wavelength to be $λ_{0} nm (= λ_{0} x 10^{- 9} m)$ , the kinetic energy of the radiation is given as:

$h (v - v_{0}) = \frac{1}{2} {mv}^{2}$

Three different equalities can be formed by the given value as:

$hc (\frac{1}{λ} - \frac{1}{λ_{0}}) = \frac{1}{2} {mv}^{2}$
$hc (\frac{1}{500 x 10^{9}} - \frac{1}{λ_{0} x 10^{- 9} m}) = \frac{1}{2} m (2.55 x 10^{+ 5} x 10^{- 2} m / s)$
$\frac{hc}{10^{- 9} m} [\frac{1}{500} - \frac{1}{λ_{0}}] = \frac{1}{2} m {(2.55 x 10^{+ 3} m / s)}^{2} (1)$

Similarly,

$\frac{hc}{10^{- 9} m} [\frac{1}{450} - \frac{1}{λ_{0}}] = \frac{1}{2} m {(3.45 x 10^{+ 3} m / s)}^{2} (2)$
$\frac{hc}{10^{- 9} m} [\frac{1}{400} - \frac{1}{λ_{0}}] = \frac{1}{2} m {(5.35 x 10^{+ 3} m / s)}^{2} (3)$

Dividing equation (3) by equation (1):

$\frac{[\frac{1}{400} - \frac{1}{λ_{0}}]}{[\frac{1}{500} - \frac{1}{λ_{0}}]} = \frac{{(5.35 x 10^{+ 3} m / s)}^{2}}{{(2.55 x 10^{+ 3} m / s)}^{2}}$
$\frac{5 λ_{0} - 2000}{4 λ_{0} - 2000} = {(\frac{5.35}{2.55})}^{2} = \frac{28.6225}{6.5025}$
$\frac{5 λ_{0} - 2000}{4 λ_{0} - 2000} = 4.40177$
$17.6070 λ_{0} - 5 λ_{0} = 8803.537 - 2000$
$λ_{0} = \frac{6805.537}{12.607}$
$λ_{0} = 539.8 nm$
$λ_{0} = 540 nm$

So, threshold wavelength $(λ_{0})$ = 540 nm

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