JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Atomic Structure

For

He^{+}

, a transition takes place from the orbit of radius 105.8 pm to the orbit of radius 26.45 pm.
The wavelength (in nm) of the emitted photon during the transition is ____.
[Use:
Bohr radius,

a = 52.9

pm
Rydberg constant,

R_{H} = 2.2 \times 1 0^{- 18} J

Planck's constant,

h = 6.6 \times 1 0^{- 34} J s

Speed of light,

c = 3 \times 1 0^{8} m s^{- 1}

]

Answer: 30

Solution

For a hydrogen-like ion, the radius of the nth orbit is:

r_{n} = \frac{n ^{2} a _{0}}{Z}

where

a_{0} = 52.9

pm and for

He^{+}

Z = 2

.

Given

r_{2} = 105.8

pm:

105.8 = \frac{n _{2}^{2} \times 52.9}{2} \Rightarrow n_{2}^{2} = 4 \Rightarrow n_{2} = 2

Given

r_{1} = 26.45

pm:

26.45 = \frac{n _{1}^{2} \times 52.9}{2} \Rightarrow n_{1}^{2} = 1 \Rightarrow n_{1} = 1

The transition is from

n = 2

n = 1

.

The energy of the emitted photon is:

Δ E = R_{H} Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}) = 2.2 \times 1 0^{- 18} \times 4 \times (1 - \frac{1}{4}) = 2.2 \times 1 0^{- 18} \times 4 \times \frac{3}{4} = 6.6 \times 1 0^{- 18} J

The wavelength is:

λ = \frac{h c}{Δ E} = \frac{6.6 \times 1 0 ^{- 34} \times 3 \times 1 0 ^{8}}{6.6 \times 1 0 ^{- 18}} = 3 \times 1 0^{- 8} m = 30 nm

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Integer answer

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JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Atomic Structure

Solution

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