Question

The maximum kinetic energy of electrons ejected from barium (whose work function is 2.50 eV) when it is illuminated by light of wavelength 350 nm is

A. 0.20 eV.

B. 0.41 eV.

C. 0.63 eV.

D. 0.95 eV.

E. 1.05 eV.

Answer 1

E. 1.05 eV

Answer 2

To calculate the maximum kinetic energy of electrons ejected from barium when illuminated by light of wavelength 350 nm, we can use the equation for the photoelectric effect:

Kinetic Energy (KE) = Energy of incident light - Work function

The energy of light can be calculated using the equation:

Energy of light (E) = (Planck's constant * speed of light) / wavelength

Given: Work function of barium (Φ) = 2.50 eV Wavelength of light (λ) = 350 nm = 350 * 10^(-9) m

Planck's constant (h) = 4.135667696 x 10^(-15) eV s Speed of light (c) = 2.998 x 10^8 m/s

Let's calculate the energy of incident light (E):

E = (h * c) / λ E = (4.135667696 x 10^(-15) eV s * 2.998 x 10^8 m/s) / (350 * 10^(-9) m) E = 11.87 eV

Now, let's calculate the maximum kinetic energy of the ejected electrons (KE):

KE = E - Φ KE = 11.87 eV - 2.50 eV KE = 9.37 eV

Answer:The maximum kinetic energy of electrons ejected from barium when illuminated by light of wavelength 350 nm is approximately 9.37 eV.

Option:B. 0.41 eV. (Incorrect)