The wavelength of light emitted in the visible region by He+ ions after collisions with H atoms is [2008]

Question

In a mixture of ${H–He}^{+}$ gas ( ${He}^{+}$ is a singly ionized He atom), H atoms and ${He}^{+}$ ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to ${He}^{+}$ ions (by collisions). Assume that the Bohr model of the atom is exactly valid. [2008]

Q. The wavelength of light emitted in the visible region by ${He}^{+}$ ions after collisions with H atoms is

Smart Achievers · Accepted Answer

(3)

Wavelength of visible light lies in the range,

$λ_{1} = 4000 Å$ to ${He}^{+}$

Therefore, $E_{1} = \frac{12375}{λ_{1}} = \frac{12375}{4000} = 3.09 eV$

$E_{2} = \frac{12375}{λ_{2}} = \frac{12375}{7000} = 1.77 eV$

For ${He}^{+}$ atom, in the transition from $n = 4$ to $n = 3$ , the energy of the photon released will lie between $E_{1}$ and $E_{2}$ .

$E_{4} - E_{3} = - 3.4 - (- 6.04) = 2.64 eV$

$∴$ Wavelength of the photon corresponding to this energy (2.64 eV),

$λ = \frac{12375}{2.64} Å = 4687.5 Å = 4.68 \times 10^{- 7} m$

Solution

1 $6.5 \times 10^{- 7} m$

2 $5.6 \times 10^{- 7} m$

3 $4.8 \times 10^{- 7} m$

4 $4.0 \times 10^{- 7} m$

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Solution

1 6.5×10-7 m

2 5.6×10-7 m

3 4.8×10-7 m

4 4.0×10-7 m

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1 $6.5 \times 10^{- 7} m$

2 $5.6 \times 10^{- 7} m$

3 $4.8 \times 10^{- 7} m$

4 $4.0 \times 10^{- 7} m$