Question

In a hydrogen atom, the energy of an electron in first Bohr’s orbit is $13.12 \times 10^{5}$ J $mol^{-1}$. What is the energy required for its excitation to Bohr’s second orbit?

Answer 1

"The expression for the energy of electron of hydrogen is

${ E }_{ n }\quad =\quad \frac { 2{\pi}^{2}{ m }_{ e }^{ 4 } }{ { n }^{ 2 }{ h }^{ 2 } }$

When, n = 1

${ E }_{ 1 }\quad =\quad \frac { 2\quad \times \quad { \left( \frac { 22 }{ 7 }\right)}^{ 2 }\quad \times \quad { \left( 9.109\quad \times \quad { 10 }^{ -31 }\right)}^{ 4 } }{ { 1 }^{ 2 }\quad \times \quad { \left( 6.626\quad \times \quad { 10 }^{ -34 }\right)}^{ 2 } } \quad =\quad -13.12\quad \times \quad { 10 }^{ 5 }\quad J{ mol }^{ -1 }$

When, n = 2

${ E }_{ 2 }\quad =\quad \frac { 2\quad \times \quad { \left( \frac { 22 }{ 7 }\right)}^{ 2 }\quad \times \quad { \left( 9.109\quad \times \quad { 10 }^{ -31 }\right)}^{ 4 } }{ { 2 }^{ 2 }\quad \times \quad { \left( 6.626\quad \times \quad { 10 }^{ -34 }\right)}^{ 2 } } \quad =\quad \frac { -13.12\quad \times \quad { 10 }^{ 5 } }{ 4 } \quad J{ mol }^{ -1 }\quad =\quad -3.28\quad \times \quad { 10 }^{ 5 }\quad J{ mol }^{ -1 }$

The energy required for the excitation:

$\Delta E\quad =\quad { E }_{ 2 }\quad -\quad { E }_{ 1 }\quad =\quad \left( -3.28\quad \times \quad { 10 }^{ 5 } \right) \quad -\quad \left( -13.12\quad \times \quad { 10 }^{ 5 } \right) \quad =\quad 9.84\quad \times \quad { 10 }^{ 5 }\quad J{ mol }^{ -1 }$"

Answer 2

Answer:

Energy in an orbit of hydrogen atom, E

n

is given by

E

n

=

n

2

h

2

2π

2

me

4

Z

2

=

n

2

constant

=

n

2

−13.6eV

For 1

st

orbit, n=1, hence E

1

=

1

−13.6eV

=−13.6eV

For 2

nd

orbit, n=2, hence E

2

=

4

−13.6eV

=−3.4eV

For 3

rd

orbit, n=3, hence E

3

=

9

−13.6eV

=−1.51eV

For 4

th

orbit, n=4, hence E

4

=

16

−13.6eV

=−0.85eV