Question

Find the value of wave number in terms
a Rydberg's constant, when transition
electron takes place between two levels
of He+ ion whose sum is the 4 and difference
is 2.​

Answer 1

Answer:

Wave no. : 32R/9

Explanation:

n2 - n1 = 4

n2 + n1=2

we get n2=3 and n1=1

Wave number = Ryd. constant x Z square x [ 1/n1^2 - 1/n2^2]

where z=2 (since it is a helium atom)

substituting the values we get W.N: 32R/9

Answer 2

The value of wave number in terms a Rydberg's constant, when transition electron takes place between two levels of He+ ion is $\frac{32}{9}\times R_H$

Explanation:

Using Rydberg's equation

$\frac{1}{\lambda}=R_H\times Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

where, $\lambda$ = wavelength

$\frac{1}{\lambda}$ = $\nu$ = wavenumber

Z = Atomic number =  2 ( for helium)

$n_1$ = Lower energy level

$n_2$ = Upper energy level

$R_H$ = Rydberg's constant

$n_2$  + $n_1$ = 4

$n_2$  - $n_1$ = 2

Thus $n_1$ = 1

$n_2$ = 3

Putting values in above equation, we get:

$\frac{1}{\lambda}=R_H\times 2^2\left(\frac{1}{1^2}-\frac{1}{9^2}\right)$

$\frac{1}{\lambda}=\frac{32}{9}\times R_H$

$\nu=\frac{32}{9}\times R_H$

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