Question

When an electron falls from (n+2) state to (n) state in a He+ ion the photon emitted has energy 6.172×10^-19 joules what is the value of n

Answer 1

An electron falls from (n+2) state to n state in a He^+ ion.

so, Z = 2 ( atomic number of Helium is Z= 2)

$n_1=(n+2)$

$n_1=n$

photon emitted has energy, $E=-2.176 \times10^{-18}Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

or, 6.172 × 10^-19 = 2.176 × 10^-18 × 4[ 1/n² - 1/(n + 2)²]

or, 0.6172/(2.176 × 4) = [ 1/n² - 1/(n + 2)² ]

or, 0.0709099= [1/n² - 1/(n + 2)² ]

or, 0.071 = [1/n² - 1/(n + 2)²]

here, 0.071 ≈ 16/225

so, 16/225 = 1/9 - 1/25

or, 1/9 - 1/25 = 1/n² - 1/(n + 2)²

or, 1/3² - 1/(3 + 2)² = 1/n² - 1/(n + 2)²

on comparing, n = 3

answer the value of n = 3

Answer 2

Answer:n=3

Explanation: