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Question : 8
Total: 27
A photon emitted during the de-excitation of electron from a state n 2 eV 0.55 V n
OR
A hydrogen atom in the ground state is excited by an electron beam of12.5 eV
OR
A hydrogen atom in the ground state is excited by an electron beam of
Solution:
● For writing Einstein's photoelectric equation
● For writing,E n = −
● For finding the value ofn
From photoelectric equation,h v = ϕ 0 + e V s
= 2 + 0.55 = 2.55 eV
Given,E n =
The energy difference
∆ E = − 3.4 − ( − 2.55 ) eV = − 0.85 eV
∴
= − 0.85
∴ n = 4
OR
Calculation of energy in excited state Formula
Finding out the maximum number of lines.
Energy in ground state,E 1 = − 13.6 eV
Energy supplied= 12.5 eV
Energy in excited state,− 13.6 + 12.5 = − 1.1 eV
But,E n =
= − 1.1
n = 3
Maximum number of lines= 3
● For writing,
● For finding the value of
From photoelectric equation,
Given,
The energy difference
OR
Calculation of energy in excited state Formula
Finding out the maximum number of lines.
Energy in ground state,
Energy supplied
Energy in excited state,
But,
Maximum number of lines
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