Question

The wavelength 6.577 x 10^-5 cm is observed in the atomic emission spectrum of hydrogen.

a. Show that this is a transition line in the Balmer Series.

b. What is the wavelength of the next higher-energy line?

Answer 1

Solution :-

a) In the balmar series the electron transition takes place from n >3 or n= 3 to tn=2

So lets calculate the wavelength for the transition from n=3 to n=2

1/ l = R[(1/nf^2)-(1/ni^2)]

1/l = 1.097*10^7   * [(1/2^2)-(1/3^2)]

1/l = 1.52*10^6

1/1.52*10^6 = l

6.56*10^-7 m = l

So the 6.56*10^-7 m * 1 *10^9 nm / 1 m = 656 nm

So this wavelength is the is the same as given in the question that is 6.56*10^-5 cm so the given transition is belong to balmer series for the n=3 to n=2 level transaction

Now part b) the higher level transition is n=4 to n=2

So lets calculate the wavelength for this transition

1/ l = R[(1/nf^2)-(1/ni^2)]

1/l = 1.097*10^7   * [(1/2^2)-(1/4^2)]

1/l= 2.06*10^6

1/ 2.06*10^6 = l

4.86*10^-7 m = l

4.86*10^-7 m * 100 cm / 1 m = 4.86*10^-5 cm

So the wavelength of the higher transition line is 4.86*10^-5 cm