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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]
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
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
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