Question

The 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

Answer 1

(a) Assuming the threshold wavelength to be λ0 nm (= λ0 × 10-9 m) , the kinetic energy of the radiation is given as: h (v - v0) = $\frac{1}{2}$mv2
Three different equalities can be formed by the given value as:
hc ($\frac{1}{\lambda}-\frac{1}{\lambda_0}$) =$\frac{1}{2}$mv2
hc ($\frac{1}{500\times10^{9}}-\frac{1}{\lambda_0\times10^{-9}m}$) = $\frac{1}{2}$m (2.55 × 10+5 × 10-2 ms-1)
$\frac{hc}{10^{-9}}m[\frac{1}{500}-\frac{1}{\lambda_0}]=\frac{1}{2}m$(2.55 × 10+3 ms-1)2 …....(1)
Similarly,$\frac{hc}{10^{-9}m}$ [$\frac{1}{450}-\frac{1}{\lambda_0}$] = $\frac{1}{2}m$ (3.45 × 10+3 ms-1)2 …...(2)
$\frac{hc}{10^{-9}m}$[$\frac{1}{400}-\frac{1}{\lambda_0}$] = $\frac{1}{2}$ m (5.35×10+3 ms-1)2 …....(3)
Dividing equation (3) by equation (1):
$\frac{[\frac{\lambda_0-400}{400\lambda_0}]}{[\frac{\lambda_0-500}{500\lambda_0}]}=\frac{(5.35\times10^{+3}ms^{-1})^2}{(2.55\times10^{+3}ms^{-1})^2}$
$\frac{5\lambda_0-2000}{4\lambda_0-2000}=(\frac{5.35}{2.55})^2=\frac{28.6225}{6.5025}$
$\frac{5\lambda_0-2000}{4\lambda_0-2000}=4.40177$
17.6070 λ0 - 5 λ0 = 8803.537 - 2000
λ0 = $\frac{ 680.537}{12.607}$
λ0 = 539.8 nm
λ0 ≈ 540 nm
∴ Threshold wavelength (λ0) = 540 nm