Home
Class 12
PHYSICS
The transition from the state n = 4 to n...

The transition from the state n = 4 to n = 3 in a hydrogen - like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition

A

`2 rarr 1`

B

`3 rarr 2`

C

`4 rarr 2`

D

`5 rarr 4`

Text Solution

Verified by Experts

The correct Answer is:
D
In the Lyman series we get energy in the UV region. In the Balmer series we get energy in the visible region. In the Paschen/Brackett/P Fund series we get energy in the IR region.
We know that = `1/lamda = R (1/n_1^2-1/n_2^2)`
for hydrogen atom fore transition from `n_2 rarr n_1`.
For 4 to 3
`E prop 1/(lamda) = R (1/9 -1/16) = 7/(9 xx16)R`
For 2 to 1, 3 to 2, and 4 to 2 we get more values of `1/lamda` , that is, more energy than `4rarr3`.
IR radiation has less energy than UV radiation.
Promotional Banner

Similar Questions

Explore conceptually related problems

The transition from the state n = 3 to n = 1 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from

In a sample of hydrogen atoms, electrons jump from 10^(th) excited state to ground state. If X are the number of different ultraviolet radiations Y are the number of different visible radiations and Z are the possible number of different infrared radiations from 10^(th) excited state to third level. The value Z-(x + y) is (Assuming all the Balmer lines lie with in visible region)

Assertion : In Li^(2+) ion, an electron make transition from higher state to n=2. Then the photon observed will fall in the visible range. Reason : Line falling in n=2 is balmer series line which belongs to visible range in all type of H-like atom orion.

Calculate the energy required to remove an electron completely from n = 2 orbit of hydrogen atom, What is the longest wavelength of light in cm that can be used to cause this transition?