Ultraviolet light of wavelengths 800 Å and 700 Å, when allowed to fall on hydrogen atom in their ground state, is able to liberate electrons with kinetic energy 1.8 eV and 4.0 eV respectively. Find the value of Planck's constant.

Solution

The energy E is related to wavelength λ, velocity of light c and Planck’s constant by the relation,

E = \frac{hc}{λ} (h \to Planck' s constant)

For radiations of wavelengths λ₁ and λ₂ respectively, energies E₁ and E₂.

Energy,

E_{1} = \frac{hc}{λ_{1}} and E_{2} = \frac{hc}{λ_{2}}

On subtracting the two,

E_{1} - E_{2} = hc (\frac{1}{λ_{1}} - \frac{1}{λ_{2}})

\Rightarrow

h = \frac{(E_{1} - E_{2}) λ_{1} λ_{2}}{c (λ_{2} - λ_{1})}

∴

h = \frac{(4.0 - 1.8) \times 1.6 \times 10^{- 19} \times 700 \times 10^{- 10} \times 800 \times 10^{- 10}}{3 \times 10^{8} \times 100 \times 10^{- 10}}

i.e.,

h = 6.57 \times 10^{- 34} Js .

is thevalue of the planck's constant.

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