CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

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 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 eV is emitted. Find n,Z and the ground state energy (in eV) for 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 13.6 eV.

A
3,4,10.5 eV.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
8,4,10.5 eV.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2,4,10.5 eV.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
2,4,15.5 eV.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 2,4,10.5 eV.
Given: E2nE1=204eV, E2nEn=40.8eV

Solution:
Let us assume that the ground state energy (in eV) be E1
Then from the given condition, we have
E2nE1=204eV or
E14n2E1=204eV or
E1(14n21)=204eV.....(i)
E2nEn=40.8eV or

E14n2E1n2=40.8eV or
E1(34n2)=40.8eV....(ii)
From Eqs. (i) and (ii), we get
114n234n2=5 or

1=14n2+154n2 or

4n2=1 or

n =2
From Eq. (ii)

E1=43n2(40.8)eV=43n2(40.8)eV or
E1=217.6eVE1=(13.6)Z2

Z2=E113.6=217.613.6=16
Z=4
Emin=E2nE2n1=E14n2E1(2n1)2
Emin=10.58eV


Hence C is the correct option

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Characteristics of Bohr's Model
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon