Which of the transitions in hydrogen atom emits a photon of lowest frequency (n = quantum number)
A)\[n= 2 to\,\,n =1\]
B)\[n =4 to\,\,n = 3\]
C)\[n= 3 \,to\,\,n = 1\]
D)\[n = 4\,\,to\,\,n = 2\]
Correct Answer:
B
Solution :
\[{{E}_{1}}\,\,=\,\,-13.6-(-\,3.4)=-10.2\,\,eV\] \[{{E}_{2}}\,\,=\,\,-13.6-(-\,1.51)=-12.09\,\,eV\] \[{{E}_{3}}\,\,=\,\,-1.51-(-\,0.85)=-0.66\,\,eV\] \[{{E}_{4}}\,\,=\,\,-3.4-(-1.51)=-1.89\,\,eV\] \[{{E}_{3}}\] is least, i.e., frequency is lowest.