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An electron in hydrogen atom first jumps...

An electron in hydrogen atom first jumps from second excited state to first excited state and then, from first excited state to ground state. Let the ratio of wavelength, momentum and energy of photons in the two cases be x,y and z, then select the correct answer/(s)

A

`z=1/x`

B

`x=9/4`

C

`y=27/5`

D

`z=5/27`

Text Solution

Verified by Experts

The correct Answer is:
A, C

`E_1= (hc)/(lamda) [1/n_1^2-1/n_2^2]=-13.6 xx3/4`
For second excited state to first excited state
`E_2 = -13.6 xx5/36`
`E_1/E_2 = (3/4)/(5/36)=27/5 =z," " 1/x = ((1)/lamda_1)/(1/lamda_2)=lamda_2/lamda_1=(R(3//4))/(R(5//36))= (36xx3)/(4xx5) = 27/5 " ":. x = 5/27`
`prop1/lamda:. p_1/p_2 =(1/lamda_1)/(1/lamda_2)=lamda_2/lamda_1=27/5=y ,x = 5/27 , y = 27/5 , z = 27/5`
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