Question

wavelength of high energy transition of H atom is 9.12 NM calculate wavelength of he atom

Answer 1

For maximum energy, n1 = 1 and n2 = ∞
1/λ = RHZ2 \left ( \frac{1}{n_{2}^{1}} - \frac{1}{n_{2}^{2}}} \right )
Since RH is a constant and transition remains the same
1/λ ∝ Z2 ; λHe/λH = \frac{2_{H}^{2}}{2_{He}^{2}}= 1/4
Hence, λHe = 1/4 * 91.2 = 22.8 nm

Answer 2

● Answer -
λ(He) = 22.8 nm

● Explaination -
Transition wavelength for H-atom is given by -
1/λ1 = RH (1/n1^2 - 1/n2^2) ...(1)

Transition wavelength for He-atom is given by -
1/λ2 = RH.Z^2 (1/n1^2 - 1/n2^2) ...(2)

Dividing (1) by (2),
λ2 / λ1 = 1 / Z^2
λ2 = λ1 / Z^2
λ2 = 91.2 / 2^2
λ2 = 22.8 nm

Therefore, wavelength of transition of He atom is 22.8 nm.


Hope this helps you...