Question

The frequency of light emitted for the transition n = 4 to n = 2 of He+ is equal to the transition in H atom corresponding to which of the following?

• A. n = 2 to n = 1
• B. n = 3 to n = 2
• C. n = 4 to n = 3
• D. n = 3 to n = 1

Answer 1

Answer: (A)

n = 2 to n = 1

Answer 2

hn = DE = 13.6 z2 $\left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)$

$\upsilon_{\text{ He+}}\text{ = }\upsilon\text{H }\text{×}\text{ }\text{z}^{2}\$

$\left( \frac{1}{\left( \frac{n_{1}}{2} \right)^{2}} - \frac{1}{\left( \frac{n_{2}}{2} \right)^{2}} \right)$

$= \upsilon H\left( \frac{1}{\left( \frac{2}{2} \right)^{2}} - \frac{1}{\left( \frac{4}{2} \right)^{2}} \right)$

For H-atom

n1 = 1, n2 = 2