Question

When an electron in hydrogen atom jumps from third excited state to the ground state, the de-Broglie wavelength associated with the electron becomes ____________.

Options

• `(2/3)^"rd"`

• half

• `(1/4)^"th"`

• `(1/3)^"rd"`

Answer 1

When an electron in hydrogen atom jumps from third excited state to the ground state, the de-Broglie wavelength associated with the electron becomes `(1/4)^"th"`.

Explanation:

`lambda = "h"/"p"      "p" = "mv"`

`"E" = 1/2 "mv"^2 = "p"^2/(2"m")`

`therefore "p" = sqrt(2"mE")`

`lambda_1/lambda_2 = "p"_2/"p"_1 = sqrt ("E"_2/"E"_1)`

`"Now  E"_"n" = 13.6/"n"^2`

`therefore  "E"prop 1/"n"^2`

For third excited state "n" = 4.

So, ratio of wavelength of ground state (n = 1) to 3rd excited state (n = 4) is

`lambda_2/lambda_1 = sqrt("E"_1/"E"_2) = sqrt(1/16) = 1/4`