A singly ionized helium atom is in the ground state. It absorbs energy and makes a transition to the n = 5 excited state. The ion returns to the ground state by emitting FOUR photons ONLY. What is the wavelength of the second highest energy photon?
Emn= (13.6Z2)((1/m2)-(1/n2)) eV
For transition of second highest energy from 5th energy level putting n= 5 and m= 2 we get
E25 = (13.6 × 42) ×21/25= 182.78 eV
Wavelength of photon = velocity of light / frequency of photon
= hc/E25 = 6.62×10-34 × 3× 108 /( 182.78 × 1.6 × 10-19)
= 0.06791 × 10-7m
= 679.1 × 10-10m
A singly ionized helium atom is in the ground state. It absorbs energy and makes a...
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