If shortest wavelength of hydrogen atom in Lyman series is x , then longest wavelength in Balmer series of He^+ is:

Question

Quantrex Academy · Accepted Answer

Correct answer: A. (9x)/(5)

For the shortest wavelength in the Lyman series of the hydrogen atom (

Z=1

), the transition is from

n_2 = \infty

to

n_1 = 1

.

\dfrac{1}{x} = R(1)^2 \left(\dfrac{1}{1^2} - \dfrac{1}{\infty^2}\right) = R

This gives

R = \dfrac{1}{x}

.

For the longest wavelength in the Balmer series of

\text{He}^+

(

Z=2

), the transition is from

n_2 = 3

to

n_1 = 2

.

\dfrac{1}{\lambda} = R(2)^2 \left(\dfrac{1}{2^2} - \dfrac{1}{3^2}\right)

\dfrac{1}{\lambda} = 4R \left(\dfrac{1}{4} - \dfrac{1}{9}\right)

\dfrac{1}{\lambda} = 4R \left(\dfrac{5}{36}\right) = \dfrac{5R}{9}

Substituting

R = \dfrac{1}{x}

:

\dfrac{1}{\lambda} = \dfrac{5}{9x}

\lambda = \dfrac{9x}{5}

Answer:

\dfrac{9x}{5}

If shortest wavelength of hydrogen atom in Lyman series is x , then...