Question

how much will the kinetic energy and total energy of an electron in H atom change if the atom emits a photon of wavelength 4860 Å

Answer 1

Use lemda = h/mv to find Velocity V
Here lemda is wavelength
M is mass of electron

Use KE=1/2m v square to find KE
Here M is mass of electron.

Hope you get your answer.

Answer 2

Hey dear,

● Answer -
∆E = ∆KE = 2.546 eV

● Explanation -
- As the threshold energy remains constant, change in energy is equal to change in kinetic energy.

Change in kinetic energy is given by -
∆E = (1/2)mv^2 = hc / λ
∆E = (6.6×10^-34 × 3×10^8) / (4.86×10^-7)
∆E = 4.074×10^-19 J
∆E = 2.546 eV

Hence, change in kinetic energy and change in total energy both is 2.546 eV.

Hope this helps you...