The Lyman series for the hydrogen atom corresponds to electronic transitions tha

Question

The Lyman series for the hydrogen atom corresponds to electronic transitions that terminate in the state with quantum number n = 1.(a) Find the longest-wavelength photon emitted in the Lyman series and determine its frequency and energy.

(b) Find the shortest-wavelength photon emitted in the same series.
nm

EXERCISEHINTS:  GETTING STARTED  |  I'M STUCK!

(a) Calculate the energy of the shortest wavelength photon emitted in the Balmer series for hydrogen.
E =  eV

(b) Calculate the wavelength of the photon emitted when an electron transits from n = 5 to n = 2.
=  nm

Explanation / Answer

(A)
1/ = Rh * (1 - 1/n^2)
Where,
Rh = 1.0986 * 10^7 m^-1
For Longest Wavelength n =2

1/ = 1.0986 * 10^7 * (1 - 1/2^2)
= 1.21556 * 10^-7 m
= 121.6 nm

velocity = frequency * wavelength
v = f *
3.0 * 10^8 = f * 121.6 * 10^-9
f = 2.46 * 10^15 Hz

E = hc/
E = (6.63 * 10^-34 * 3.0*10^8)/(121.6 * 10^-7) J
E = 1.635 * 10^-20 J
E = 10.22 eV

(B)
The Lyman limit is the short-wavelength end of the hydrogen Lyman series, = 91.2 nm
It corresponds to the energy required for an electron in the hydrogen ground state to escape.

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