1 answer

Use the Bohr model to find the second longest wavelength of light in the Balmer series...

Question:

Use the Bohr model to find the second longest wavelength of light in the Balmer series for a doubly-ionized Li atom (Z = 3). Recall that the Balmer series corresponds to transitions to the n = 2 level.

Group of answer choices

41.1 nm

117 nm

73.0 nm

54.1 nm

209 nm


Answers

Longer the wavelength, smaller should be the energy difference.

for the longest wavelength of Barmer series transition should be from n=3 to n=2

for the second-longest wavelength of Barmer series transition should be from n=4 to n=2

\frac{1}{\lambda} = R Z^2 (\frac{1}{n_f ^2}- \frac{1}{n_i ^2})

= (1.097 x 107 m-1) x 32 x  (\frac{1}{2^2}- \frac{1}{4 ^2})

= 1.851 x 107 m-1

\lambda= 54.1 x 10-9 m

= 54.1 nm

.

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