JEE Advanced 2023 — Paper 2 — CHEM-III solution

JEE Advanced2023Paper 2CHEM-III

For $H e^{+}$ , 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 $^{- 1}$ ]

Your answer

Solution

The Bohr orbit radius for a one-electron ion is $r_{n} = \frac{a n ^{2}}{Z}$ with $Z = 2$ for $H e^{+}$ .

$105.8 = \frac{52.9 n _{2}^{2}}{2} \Rightarrow n_{2} = 2$ .

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

So the transition is $n = 2 \to n = 1$ . Since $R_{H}$ is given as an energy ( $2.2 \times 1 0^{- 18}$ J),

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

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

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