When an electron in hydrogen atom jumps from the third excited state to the first excited state?

Text Solution

`7//5``27//20``27//5``20//7`

Answer : C

Solution : We know, `(1)/(lambda) = R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` <br> Where `R` is the Rydberg constant <br> For case `(1)n_(1) = 3` and `n_(2) = 2` <br> `(1)/(lambda_(1)) = R((1)/(3^(2)) - (1)/(2^(2))) =- (5R)/(36)` <br> `rArr lambda_(1) =- (36)/(5R)……...(i)` <br> For case (II)`n_(1) = 2` and `n_(2) =1` <br> `(1)/(lambda_(2)) = R ((1)/(2^(2)) - (1)/(1^(2))) =- (3R)/(4)` <br> `rArr lambda_(2) =- (4)/(3R)` .......(i) <br> From (i) and (ii), `(lambda_(1))/(lambda_(2)) = (-(36)/(5R))/(-(4)/(3R)) = (27)/(5)`

An electron in a hydrogen atom first jumps from third excited state to second excited state and then from second excited state to first excited state. The ratio of wavelengths λ1:λ2 emitted in the two cases is

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When an electron in hydrogen atom jumps from the third excited state to the ground state, how would the de Broglie wave length associated with the electron change? Justify your answer.

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