Question

The work function for metallic caesium is 2.14 eV. Calculate the

Answer 1

Answer:

Part 1) No electrons are emitted

Part 2) Energy equals 2.837eV and velocity equals $0.998\times 10^{6}m/s$

Explanation:

From the basic equation of photoelectric effect we have

$h\nu =\phi +\frac{1}{2}mv^{2}$

where

$h\nu =\frac{hc}{\lambda }$ is the energy of the incident radiation

$\phi$ is the work function of the metal

$\frac{1}{2}mv^{2}$ is the kinetic energy of the emitted electrons

Thus applying values we get

1) for wavelength of 750 nanometers we have

The energy of incident radiation equals $6.636\times 10^{-34}\times\frac{3\times 10^{8}}{750\times 10^{-9}\times 1.6\times 10^{-19}}=1.66eV$

Since the energy of incident radiation is lesser than the work function of the metal hence no electron's will be emitted.

Part 2)

$K.E=6.63\times 10^{-34}\times \frac{3\times 10^{8}}{250\times 10^{-9}}-2.14\times 1.6\times 10^{-19}$

Thus kinetic energy equals $2.837eV$

Thus speed from energy is calculated as

$\\\\v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2\times 2.837\times 1.6\times 10^{-19}}{9.1\times 10^{-31}}}$

$v_{electron}=0.9988\times 10^{6}m/s$

Answer 2

Answer:a) -7.73×10^-20J

b) 3.201×10^-19J

Explanation:Kinetic energy of an electron is given by the equation: 1/2MeV^2=hv-phi(work function)

E=hc/wavelength- phi(work function)

WhereMe=mass of electron

h=plank's constant

V=velocity of ejected electron

Phi=work function=2.14ev

a) Wavelength=750nm

Substituting,

(6.636×10^-34js)(3×10^8m/s)/750×10^-9 -(2.14×1.602×10^-19/1ev

1.991×10^-25/760×10^-9 -3.428×10^-19

K.E=-7.73×10^-20J

b. Wavrlength=250nm

(6.636×10^-34)(3×10^8)/250×10^-9 -2.14×1.602×10^-19

K.E= 6.637×10^-19 -3.4283×10^-19

K.E=3.201×10^-19