Modern Physics 1 Question 51

47. A hydrogen like atom of atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204eV. If it makes a transition to quantum state n, a photon of energy 40.8eV is emitted. Find n,Z and the ground state energy (in eV ) of this atom. Also, calculate the minimum energy (in eV ) that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is 13.6eV.

(2000,6 M)

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Answer:

Correct Answer: 47. n=2,Z=4,217.6eV,10.58eV

Solution:

  1. Let ground state energy (in eV ) be E1.

Then, from the given condition

E2nE1=204eV or E14n2E1=204eV or E114n21=204eV and E2nEn=40.8eV or E14n2E1n2=40.8eV or E134n2=40.8eV

From Eqs. (i) and (ii),

114n234n2=5 or 1=14n2+154n2 or 4n2=1 or n=2

From Eq. (ii),

E1=43n2(40.8)eV=43(2)2(40.8)eVE1=217.6eVE1=(13.6)Z2Z2=E113.6=217.613.6=16ZEmin=E2nE2n1=E14n2E1(2n1)2=E116E19=7144E1=7144(217.6)eVEmin=10.58eV



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