JEE MAIN - Physics (2019 - 8th April Morning Slot - No. 6)

Radiation coming from transitions n = 2 to n = 1 of hydrogen atoms fall on He+ ions in n = 1 and n = 2 states. The possible transition of helium ions as they absorb energy from the radiation is :
n = 1 $$ \to $$ n = 4
n = 2 $$ \to $$ n = 5
n = 2 $$ \to $$ n = 4
n = 2 $$ \to $$ n = 3

Explanation

Energy released for tension n = 2 to n = 1 of hydrogen atom

$$E = 13.6{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$$

Z = 1, n1 = 1, n2 = 2

$$E = 13.6 \times 1 \times \left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)$$

$$E = 13.6 \times {3 \over 4}eV$$ = 10.2 eV

For He+ ion z = 2

(A) n = 1 to n = 4
   $$E = 13.6 \times {2^2} \times \left( {{1 \over {{1^2}}} - {1 \over {{4^2}}}} \right) = 13.6 \times {{15} \over 4}eV$$

(B) n = 2 to n = 4
   $$E = 13.6 \times {2^2} \times \left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right) = 13.6 \times {{3} \over 4}eV$$

(C) n = 2 to n = 5
   $$E = 13.6 \times {2^2} \times \left( {{1 \over {{2^2}}} - {1 \over {{5^2}}}} \right) = 13.6 \times {{21} \over 25}eV$$

(D) n = 2 to n = 3
   $$E = 13.6 \times {2^2} \times \left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right) = 13.6 \times {{5} \over 9}eV$$

So, possible transition is n = 2 $$\to$$ n = 4

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