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faust18 [17]
3 years ago
8

White light, with frequencies ranging from 4.00 x 10^14 Hz to 7.90 x 10^14 Hz, is incident on a barium surface. Given that the w

ork function of barium is 2.52 eV, find:
(a) the maximum kinetic energy of electrons ejected from this surface.
(b) the range of frequencies for which no electrons are ejected.
Physics
1 answer:
REY [17]3 years ago
5 0

Answer:

0.7515875 eV

4\times 10^{14}\leq f

Explanation:

f = Maximum frequency = 7.9\times 10^{14}\ Hz

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

W = Work function = 2.52 eV

Converting to Joules

W=2.52\times 1.6\times 10^{-19}\\\Rightarrow W=4.032\times 10^{-19}\ J

Maximum photon energy is given by

E=hf\\\Rightarrow E=6.626\times 10^{-34}\times 7.9\times 10^{14}\\\Rightarrow E=5.23454\times 10^{-19}\ J

Maximum Kinetic energy is given by

K=E-W\\\Rightarrow K=5.23454\times 10^{-19}-4.032\times 10^{-19}\\\Rightarrow K=1.20254\times 10^{-19}\ J

Converting to eV

1.20254\times 10^{-19}\times \frac{1}{1.6\times 10^{-19}}=0.7515875\ eV

The maximum kinetic energy of electrons ejected from this surface is 0.7515875 eV

W=hf\\\Rightarrow f=\frac{W}{h}\\\Rightarrow f=\frac{4.032\times 10^{-19}}{6.626\times 10^{-34}}\\\Rightarrow f=6.08512\times 10^{14}\ Hz

The range of frequencies for which no electrons are ejected is

4\times 10^{14}\leq f

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An electron has a kinetic energy of 3.00 ev. find its wavelength. (b) what if? a photon has energy 3.00 ev. find its wavelength.
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(b) By using the same procedure of part (a), we can convert the photon energy into Joules:
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\lambda= \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{7.27 \cdot 10^{14} Hz} =4.13 \cdot 10^{-7} m
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