Question

Which of the following transition will have highest emission frequency?

पर्याय

• n = 2 to n = 1

• n = 1 to n = 3

• n = 2 to n = 5

• n = 5 to u = 2

Answer 1

n = 2 to n = 1

Explanation:

To obtain emission frequency the e lectron must transit from higher to lower energy level. Hence, transitions from n = 1 to n = 3 and n = 2 to n = 5 does not contribute to emission of radiations.

`Delta "E" = "hv"`

`Rightarrow Delta "E" prop "v"`      ....(i)

`"Also," Delta "E"_"n" = "E"_1 (1/"n"_"i"^2 - 1/"n"_"f"^2)`

For transition n = 2 to n = 1

`therefore Delta "E"_1 = "E"_0 abs (1/4 - 1/1) = (3"E"_0)/4`

For transition n = 5 to n = 2

`Delta "E"_2 = "E"_0 abs(1/5^2 - 1/2^2)`

` = "E"_0 abs(1/25 - 1/4)`

` = (2  "IE"_0)/100`

`therefore Delta "E"_1 > Delta "E"_2`

`Rightarrow "v"_1 > "v"_2`     .....[From (i)]

Hence, for transition from n = 2 to n = 1 the frequency emitted will be highest.