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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)

Amit Saxena , 11 Years ago
Grade upto college level
anser 1 Answers
Navjyot Kalra

Last Activity: 11 Years ago

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 / λ = √2E0 / E0 ⇒ λ1 / λ2 = √2
Thanks
Navjot Kalra
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