The work function for metallic rubidium is 2.09 eV. Calculate the kinetic energy and the speed of the electron ejected by light of wavelength (i) 650 nm and (ii) 195 nm.

i) No electron is emitted. Hence, no kinetic energy or velocity of emitted electron

ii) KE = 6.85 x 10⁻¹⁹ J; v = 1.23 x 10⁶ m/s

Explanation:

Using the formula: Kinetic energy, KE = hc/λ - E₀

where h = the Plank constant 6.63 x 10⁻³⁴ J s,

c is velocity of light = 3.00 x 10⁸ m/s

λ = wavelength of incident light in nm

E₀ = work function in Joules

Note: 1 eV = 1.602×10⁻¹⁹ Joules

Also, using KE = 1/2 mv²

v = √(2KE/m)

where m is mass of electron = 9.11 x 10⁻³¹ Kg

i) KE = hc/λ - E₀

KE = (6.63 x 10⁻³⁴ Js x 3.00 x 10⁸ m/s) / 650 x 10⁻⁹ m - (2.09 x 1.602×10⁻¹⁹ J)

KE = 3.06 x 10⁻¹⁹ J - 3.35 x 10⁻¹⁹  = - 2.9 x 10⁻²⁰ J

Since KE is negative, no electron is emitted. Thus, there is no kinetic energy or velocity of electron.

ii) KE = hc/λ - E₀

KE = (6.63 x 10⁻³⁴ Js x 3.00 x 10⁸ m/s) / 195 x 10⁻⁹ m - (2.09 x 1.602×10⁻¹⁹ J)

KE = 10.2 x 10⁻¹⁹ J - 3.35 x 10⁻¹⁹

KE = 6.85 x 10⁻¹⁹ J

Velocity, v = √(2KE/m)

v = √( 2 x 6.85 x 10⁻¹⁹ J / 9.11 x 10⁻³¹ Kg)

v = 1.23 x 10⁶ m/s

it can be icreased by increasing pressure.

the answer is c i took the test