Question

energy of the second orbit of hydrogen is equal to the energy of:
A) 4th orbit of He+
B) 4th orbit of Li2+
C) 2nd orbit of Het
D) 2nd orbit of Li2+
​

Answer 1

Answer:a)4th orbit of He+

Explanation:

n=2

Z=1

energy of hydrogen=-13.6*Z^2/n^2

                                 =-13.6*1/4

A)n=4

   Z=2

                   He+=-1.36*4/16

                         =-1.36*1/4 eV/atom

hence,proved

Answer 2

The energy of the second orbit of hydrogen is equal to the energy of 4th orbit of $He^{+}$

Explanation:

Energy of the nth orbit by Bohr was given by:

$E_n=R_H\times \frac{Z^2}{n^2}eV$

where,

$E_n$ = energy of $n^{th}$ orbit

$R_H$ = Rydberg constant

n = number of orbit  

Z = atomic number

Energy of the second orbit of hydrogen:

$E_2=R_H\times \frac{1^2}{2^2}eV$

$E_2=\frac{R_H}{4}eV$

a) 4th orbit of $He^{+}$

$E_4=R_H\times \frac{2^2}{4^2}eV$

$E_4=\frac{R_H}{4}eV$

b) 4th orbit of $Li^{2+}$

$E_4=R_H\times \frac{3^2}{4^2}eV$

$E_4=\frac{9R_H}{16}eV$

c)  2nd orbit of $He^{+}$

$E_2=R_H\times \frac{2^2}{2^2}eV$

$E_2=R_HeV$

d) 2nd orbit of $Li^{2+}$

$E_2=R_H\times \frac{3^2}{2^2}eV$

$E_2=\frac{9R_H}{16}eV$

Thus energy of the second orbit of hydrogen is equal to the energy of 4th orbit of $He^{+}$

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