Question

26)​Calculate the longest wavelength of light that will be needed to remove an electron from the third orbit of He+ ion.

Answer 1

1/λ = RZ^2 * [1/n1^2 - 1/n2^2]
n1 = 3 and for longest wavelength n2 = 4

1/λ = R * 4 * (1/9 - 1/16)
1/λ = 28R/144
λ = 144/(28R)
λ = 36/(7R)

Wavelength required is 36/(7R)

Answer 2

we have to calculate longest wavelength of light that will be needed to remove an electron from 3rd orbit of He+ ion.

here, Atomic number of atom, Z = 2

[ atomic number of He is 2]

$n_1=3$ and for longest wavelength $n_2=4$

using Rydberg's equation,

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

putting Z = 2, $n_1=3$ and $n_2=4$

so, $\frac{1}{\lambda}$ = R(2)² [1/3² - 1/4² ]

= 4R [ 1/9 - 1/16 ]

= 4R [ 7/144]

= 7R/36

so, wavelength, $\lambda$ = 36/7R

we know, Rydberg's constant, R = 1.09677 × 10^7/m

so, longest wavelength = 36/7(1.09677 × 10^7) m

≈ 4.69 × 10^-7 m

≈ 469 nm