The photons from the Balmer series in Hydrogen spectrum having wavelength between 450 nm to 700 nm are incident on a metal surface of work function 2 eV. Find the maximum kinetic energy of ejected electron. (Given hc = 1242 eV nm)

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Navjyot Kalra · Answer

Hello Student,

Please find the answer to your question

The de Broglie wave length is given by

λ = h / mv ⇒ λ = h / √2mK

Case (i) 0 ≤ x ≤ 1

For this, potential energy is E₀ (given)

Total energy = 2E₀ (given)

∴ Kinetic energy = 2 E₀ – E₀ = E₀

λ₁= h / √2mE₀ …(i)

Case (ii) x > 1

For this, potential energy = 0 (given)

Here also total energy = 2E₀ (given)

∴ Kinetic energy = 2E₀

∴ λ₂ = h / √2m (2E₀) …(ii)

Diving (i) and (ii)

λ₁ / λ₂ = √2E₀ / E₀ ⇒ λ₁ / λ₂ = √2

Thanks
Navjot Kalra
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The photons from the Balmer series in Hydrogen spectrum having wavelength between 450 nm to 700 nm are incident on a metal surface of work function 2 eV. Find the maximum kinetic energy of ejected electron. (Given hc = 1242 eV nm)

The photons from the Balmer series in Hydrogen spectrum having wavelength between 450 nm to 700 nm are incident on a metal surface of work function 2 eV. Find the maximum kinetic energy of ejected electron. (Given hc = 1242 eV nm)

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