Question

When an electron makes a transition from
(n + 1) state to nth state, the frequency of
emitted radiations is related to n according
to (n >> 1):​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

SOLUTION :-

$\tt \dfrac{1}{λ} = RZ ^{2} ( \dfrac{1}{n \dfrac{2}{1} } - \dfrac{1}{n \dfrac{2}{2} }), \: where \: n{1} = n,n2 = n + 1$

★$\tt \dfrac{1}{λ} = RZ ^{2} ( \dfrac{1}{n ^{2} } - \dfrac{1}{(n + 1) ^{2} })$

$→ \tt \dfrac{1}{λ} = ( \dfrac{2n + 1}{n ^{2}(n + 1)^{2} }) RZ ^{2}$

$</p><p> \tt Since, n>>1;</p><p></p><p>$

$\tt Therefore, 2n+1≈2n;</p><p></p><p>$

$\tt and (n+1)2≈n2$

$★ \tt \dfrac{1}{λ} = RZ ^{2} ( \dfrac{2n}{n^{2} .n ^{2} })$

$→ \tt \dfrac{v}{c} = \dfrac{2RZ ^{2} }{ {n}^{3} } \: or \: v \: = \dfrac{2cRZ ^{2} }{ {n}^{3} }$

$→ \tt answer = \dfrac{2cRZ ^{2} }{ {n}^{3} }$