Question

An atom of He+ has its electron in the n = 3 energy level. The atom is struck by a photon with an energy of 7.27×10–19 J, exciting the electron to a higher energy level. What level has the electron been excited to?

(a) 2

(b) 4

(c) 5

(d) 6

(e) 7

Answer 1

From Rydburg's Equation , 1/ λ = R[(1/n_i²) – (1/n_f²)]
   Where R = Rydburg's constant = 10.96 x10⁶ m^-1
               λ = wavelength = ?
               n_i = initial n value = 3
               n_f = final n value = ?

Also E = hc/λ = hcR[(1/n_i²) – (1/n_f²)]

Here

h = Planck's constant = 6.626x10-34 Js

c = speed of photon(speed of light) = 3.0x108 m/s

E = energy of photon = 7.7x10^-19 J

Plug the values we get

[(1/3²) – (1/n_f²)] = E / (hcR) = 0.334

1/n_f² = -0.223

n_f² = 4

n_f = 2