Question

If the ionization energy of hydrogen is 313.8K cal per mole, then the energy of the electron in $2^{nd}$ excited state will be:-

• A. -113.2Kcal/mole
• B. -78.45Kcal/mole
• C. -313.8Kcal/mole
• D. -35Kcal/mole

Answer 1

The correct option is D -35Kcal/mole

Ionisation energy is the energy required to remove the electron from respective orbit to n = ∞ i.e. Energy required to send the electron from respective orbit n to n = ∞

Given that the energy required to remove the electron from n = 1 to n = ∞ is -313.8 kcal/mole.

we know the relation E = hc/λ

⇒ Energy required to remove the electron is inversely proportional to wavelength.

⇒ Energy required to remove the electron is Directly proportional to wavenumber.

we know that wavenumber is given by R × (1/n₁² - 1/n₂²)

given ionisation energy is when n = 1 and n = ∞

then wavenumber corresponding to this R × (1 - 0) = R

we know that in the second excited state the electron is in the orbit - n =3

and ionisation energy is the energy required to send the electron from n = 3 to n = ∞

wavenumber corresponding to this is R ×(1/3² - 1/∞²) = R/9

the wavenumber corresponding to n = 3 to n = ∞ is one ninth the wavenumber corresponding to n = 1 to n = ∞

as the energy is directly proportional to wavenumber, the energy required to send away the electron from n = 3 to n = ∞ is one ninth the energy required to send away the electron from n = 1 to n = ∞

Given that the energy required to remove the electron from n = 1 to n = ∞ is -313.8 kcal/mole.

⇒ Energy required to send away the electron from n = 3 to n = ∞ is 1/9 × -313.8

⇒ Energy required to send away the electron from n = 3 to n = ∞ is -34.87 kcal/mole.

Ionisation energy of the electron in 2nd excited state is -34.87 kcal/mole