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Gamma Decay

The value of ...

Question

The value of $(n_{1} + n_{2})$ and $(n_{2}^{2} - n_{1}^{2})$ for ${H e}^{+}$ ion in atomic spectrum are $4$ and $8$ respectively. The wavelength of emitted radiation when electrons jump from $n_{2}$ to $n_{1}$ is:

\frac{32}{9 R_{H}}

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\frac{9}{32 R_{H}}

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\frac{32}{9} R_{H}

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\frac{9}{32} R_{H}

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Solution

The correct option is B $\frac{9}{32 R_{H}}$
$E x p l a n a t i o n :$
Given:
The wavelength of the emitted photon when an electron from $n_{2}$ to $n_{1} = ?$

$(n_{2} + n_{1}) = 4 \to (1)$

$(n_{2}^{2} - n_{1}^{2}) = 8 \to (2)$
From equation (1)and (2):

$∵ (a^{2} - b^{2}) = (a + b) (a - b)$

$(n_{2}^{2} - n_{1}^{2}) = (n_{2} + n_{1}) (n_{2} - n_{1})$

$(n_{2} + n_{1}) (n_{2} - n_{1}) = 8$

$4 (n_{2} - n_{1}) = 8$

$(n_{2} - n_{1}) = 2 \to (3)$

Now add equation (2) and (3) we get

$n_{2} + n_{1} + n_{2} - n_{1} = 2 + 4$

$2 n_{2} = 6 \Rightarrow n_{2} = 3$

$n_{1} = 4 - n_{2}$

$n_{1} = 4 - 3 = 1$

$∴ n_{1} = 1, n_{2} = 3$

From $R y d b e r g^{'} s e q u a t i o n$ we have the formula for wavelength as

$\frac{1}{λ} = R_{H} Z^{2} [\frac{1}{n_{1}^{2}} \frac{1}{n_{2}^{2}}]$

where,

$λ = w a v e l e n g t h = ?$

$Z =$ atomic number $= 2$

$\Rightarrow \frac{1}{λ} = R_{H} \times 4 [\frac{1}{1} - \frac{1}{9}] = R \times 4 [\frac{9 - 1}{9}] = R_{H} \times 4 [\frac{8}{9}] = \frac{32 R_{H}}{9}$

$\Rightarrow \frac{1}{λ} = \frac{32 R_{H}}{9}$

$∴ λ = \frac{9}{32 R_{H}}$

$H e n c e t h e c o r r e c t a n s w e r i s o p t i o n (B) .$

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The value of (n1+n2) and (n22−n21) for He+ ion in atomic spectrum are 4 and 8 respectively. The wavelength of emitted radiation when electrons jump from n2 to n1 is:

The value of $(n_{1} + n_{2})$ and $(n_{2}^{2} - n_{1}^{2})$ for ${H e}^{+}$ ion in atomic spectrum are $4$ and $8$ respectively. The wavelength of emitted radiation when electrons jump from $n_{2}$ to $n_{1}$ is: