Answer:
Part 1) No electrons are emitted
Part 2) Energy equals 2.837eV and velocity equals [tex]0.998\times 10^{6}m/s[/tex]
Explanation:
From the basic equation of photoelectric effect we have
[tex]h\nu =\phi +\frac{1}{2}mv^{2}[/tex]
where
[tex]h\nu =\frac{hc}{\lambda }[/tex] is the energy of the incident radiation
[tex]\phi [/tex] is the work function of the metal
[tex]\frac{1}{2}mv^{2}[/tex] 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 [tex]6.636\times 10^{-34}\times\frac{3\times 10^{8}}{750\times 10^{-9}\times 1.6\times 10^{-19}}=1.66eV[/tex]
Since the energy of incident radiation is lesser than the work function of the metal hence no electron's will be emitted.
Part 2)
[tex]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}[/tex]
Thus kinetic energy equals [tex]2.837eV[/tex]
Thus speed from energy is calculated as
[tex]\\\\v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2\times 2.837\times 1.6\times 10^{-19}}{9.1\times 10^{-31}}}[/tex]
[tex]v_{electron}=0.9988\times 10^{6}m/s[/tex]
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
