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