Wavelength of photon absorbed, $$\lambda$$ = 330 nm = 330 $$\times$$ 10^$$-$$9 m

Planck's constant, h = 6.6 $$\times$$ 10^$$-$$34 J s

Molar mass of He, M = 4 g mol^$$-$$1 = 4 $$\times$$ 10^$$-$$3 kg mol^$$-$$1

Avogadro number, N_A = 6 $$\times$$ 10²³ mol^$$-$$1

Mass of one atom of He, $$m = {M \over {{N_A}}}$$

$$ = {{4 \times {{10}^{ - 3}}} \over {6 \times {{10}^{23}}}} = {2 \over 3} \times {10^{ - 26}}$$ kg

Velocity, = V cm/s.

Using de-Broglie equation,

$$\lambda = {h \over {mv}}$$

$$ \Rightarrow v = {h \over {m\lambda }} = {{6.6 \times {{10}^{ - 34}}} \over {2/3 \times {{10}^{ - 26}} \times 330 \times {{10}^{ - 9}}}}$$

$$ = {{6.6 \times 3 \times {{10}^{ - 34}} \times {{10}^{35}}} \over {2 \times 330}} = 0.03 \times 10 = 0.3$$ m/s

= 30 cm/s

Given that, electrons and nucleus are at infinite distance, so potential energy of H-atom is taken as zero.

Therefore, according to Bohr's model, potential energy of a H-atom with electron in its ground state = $$-$$27.2 eV

At d = d₀, nucleus-nucleus and electron-electron repulsion is absent.

Hence, potential energy will be calculated for 2 H atoms = $$-$$2 $$ \times $$ 27.2 eV = $$-$$54.4 eV

Potential energy of 1 mol H atoms in kJ

= $${{54.4 \times 6.02 \times {{10}^{23}} \times 1.6 \times {{10}^{ - 19}}} \over {1000}}$$

= $$ - 5242.4192$$ kJ/mol

As n = 4, so l = 0, 1, 2, 3 which implies the orbitals are 4s, 4p, 4d and 4f.

Now, |m_l | = 1 implies m_l = +1 and −1. Therefore, l can be 3, 2, 1, as for l = 0, m_l = 0.

For l = 1, m_l = −1, 0, +1

For l = 2, m_l = −2, −1, 0, +1, +2

For l = 3, m_l = −3, −2, −1, 0, +1, +2, +3

Therefore, there are six possible orbitals with |m_l | = 1 . Also, given that m_s = −1/2 so six electrons are possible with m_s = −1/2.