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Grade: 11
        Electron in sample of hydrogen atoms make transition from state n=x to some lower excited state. Emission spectrum from sample is found to contain only lines belonging to particular series. If one of photons had an energy of 0.6537ev. Find value of x [0.6537ev=3/4*0.85ev]
8 years ago

Answers : (1)

vikas askiitian expert
509 Points
							

E = 13.6(1/n2 - 1/x2)

E is energy os photon emmited when electron falls from higher state x to lower state n...

 

E = 0.6537ev           (given)

so , 0.6537 = 13.6(1/n2-1/x2)

 (3/4) (0.85)/13.6 = [ 1/n2 - 1/x2 ]

(3/4)/16 = 3/64 = [ 1/n2 - 1/x2 ]

now , here we can use hit & trial method , for n = 4 & x = 8 this relation is satisfied

[1/16 - 1/64] = [ 1/n2 - 1/x2 ]

n = 4 , x = 8

8 years ago
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