Question

Wave number of limiting line of Lyman series for He+ is
a) R
b) R/4
c)2R
d)4R

Answer 1

Answer:

R

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Answer 2

Answer:

The wavenumber of the limiting line of the Lyman series for $He^{+}$ calculated is equal to R.

Therefore, option a) R is correct.

Explanation:

Given,

The Lyman series for $He^{+}$, i.e., n₁ = $1$

And for the limiting line, n₂ = $\infty$

Here, n₁ and n₂ are integers or principal quantum numbers.

The wavenumber of the limiting line of the Lyman series for $He^{+}$ =?

As we know,

• Wavenumber is the reciprocal of wavelength ( λ ).

• Wavenumber = $\frac{1}{\lambda }$

Also,

• $\frac{1}{\lambda }$ = $R [\frac{1}{n_{1}^{2} } - \frac{1}{n_{2}^{2} } ]$

Here, R = Raydberg's constant

After putting the values of the integers ( n₁ and n₂ ) in the equation, we get:

• $\frac{1}{\lambda }$ = $R [\frac{1}{1^{2} } - \frac{1}{\infty^{2} } ]$         ( $\frac{1}{\infty^{2} } = 0$, because it is a very, very small quantity )
• $\frac{1}{\lambda }$ = $R [\frac{1}{1 } - 0 ]$
• $\frac{1}{\lambda }$ = $R$

Hence, the wavenumber of the limiting line of the Lyman series for $He^{+}$ calculated is equal to R.