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