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The ground state energy of hydrogen atom is -13.6 eV. The photon emitted during the transition of electron from n = 1 to n = 1 unknown work function. - Physics

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प्रश्न

The ground state energy of hydrogen atoms is -13.6 eV. The photon emitted during the transition of electron from n = 3 to n = 1 unknown work function. The photoelectrons are emitted from the material with a maximum kinetic energy of 9 eV. Calculate the threshold wavelength of the material used.

बेरीज

उत्तर

For a transition from n = 3 to n = 1 state, the energy of the emitted photon,

hv = E2 – E1 = `13.6[1/1^2 - 1/3^2]` eV = 12.1 eV.

From Einstein’s photoelectric equation,

hv + Kmax + W0

∴ W0 = hv – Kmax = 12.1 – 9 = 3.1 eV

Threshold wavelength,

λth = `("hc")/"W"_0`

= `(6.62 xx 10^-34 xx 3 xx 10^8)/(3.1 xx 1.6 xx 10^-19)`

= 4 × 10–7m

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