Radiation coming from transition $\mathrm{n}=2$ to $\mathrm{n}=1$ of hydrogen atoms fall of $\mathrm{He}^{+}$ions in $\mathrm{n}=$ 1 and $\mathrm{n}=2$ states. The possible. Transition of helium ions as they absorb energy from the radiation is:

(A) $\mathrm{n}=2 \rightarrow \mathrm{n}=4$

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Modern Physics

Radiation coming from transition

n = 2

n = 1

of hydrogen atoms fall of

He^{+}

ions in

n =

1 and

n = 2

states. The possible. Transition of helium ions as they absorb energy from the radiation is:

(A) $n = 2 \to n = 4$ ✓ Correct
(B) $n = 2 \to n = 5$
(C) $n = 2 \to n = 3$
(D) $n = 1 \to n = 4$

Answer: (A)

Solution

The photon from hydrogen's

n = 2 \to 1

drop carries

E = 13.6 (1 - \frac{1}{4}) = 13.6 \times \frac{3}{4}

eV. A

He^{+}

ion has

Z = 2

, so its levels are

E_{n} = - 13.6 \frac{4}{n ^{2}}

eV. Testing the jump from

n = 2

n = 4

, the energy absorbed is

13.6 \times 4 (\frac{1}{4} - \frac{1}{16}) = 13.6 \times 4 \times \frac{3}{16} = 13.6 \times \frac{3}{4}

eV, exactly the incoming photon. The helium ion is therefore raised from

n = 2

n = 4

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option

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