Photons of equal energy were incident on two different gas samples. One sample containing H-atoms in the ground state and the other sample containing H-atoms in some excited state with a principal quantum number ‘n’. The photonic beams totally ionise the H-atoms. If the difference in the kinetic energy of the ejected electrons in the two different cases is 12.75 eV. Then find the principal quantum number ‘n’ of the excited state.
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a
1
b
2
c
3
d
4
answer is D.
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Detailed Solution
En=-K.EK.E2-K.E1=E1-E2=13.6Z21n12-1n2213.6 Z2 112 − 1n2= 12.75 given.∴ n2=16 or n=4.
Photons of equal energy were incident on two different gas samples. One sample containing H-atoms in the ground state and the other sample containing H-atoms in some excited state with a principal quantum number ‘n’. The photonic beams totally ionise the H-atoms. If the difference in the kinetic energy of the ejected electrons in the two different cases is 12.75 eV. Then find the principal quantum number ‘n’ of the excited state.