Energy E of a hydrogen atom with principal quantum number

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Energy E of a hydrogen atom with principal quantum number n is given by E = – 13.6/n² eV. The energy of photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen is approximately

1. 1.9 eV.
2. 1.5 eV
3. 0.85 eV
4. 3.4 eV

Correct Option: A

∆E = E₃ – E₂

=	- 13.6	+	13.6
	3²		2²

= 13.6		1	-	1		eV = 1.9 eV
		4		9