Home Page Textbooks Physics Page 861: Practice Problems

Physics

1st Edition

Walker

ISBN: 9780133256925

Textbook solutions

All Solutions

Page 861: Practice Problems

Exercise 9

Solution 1

Solution 2

Step 1

1 of 2

$textbf{Given: }$

$f_o = 4.2 times 10^{15}$ Hz

$textbf{Find: }$ Work function, $W_o$

We calculate for $W_o$ as follows:

$$
begin{align*}
f_o &= dfrac{W_o}{h} \
W_o &= f_o times h \
&= (4.2 times 10^{15} text{ Hz}) times (6.626 times 10^{-34} text{ Js}) \
&= 2.8 times 10^{-18} text{ J} times left(dfrac{1 text{eV}}{1.6 times 10^{-19} text{ J}}right) \
&= boxed{17.4 text{ eV}}
end{align*}
$$

Result

2 of 2

$$
W_o = 17.4 text{ eV}
$$

Step 1

1 of 2

The cutoff frequency is the lowest frequency that causes photoelectric effect so here the kinetic energy of ejected electron is zero and we write

$$
W_0 = hnu = 2.8times 10^{-18}text{ J}.
$$

Result

2 of 2

Click here for the solution.

Exercise 10

Solution 1

Solution 2

Step 1

1 of 3

$textbf{Given: }$

$f_1 = 1.0 times 10^{15}$ Hz

$f_2 = 2.0 times 10^{15}$ Hz

$W_o = 4.58$ eV

To be able to determine if the photon can eject electrons from gold, it must have an energy greater than the work function or $E > W_o$

a) We calculate the energy carried by this photon as follows:

$$
begin{align*}
E_1 &= hf_1 \
&= (6.626 times 10^{-34} text{ Js}) times (1.0 times 10^{15} text{ Hz}) \
&= 6.626 times 10^{-19} text{J} times left(dfrac{1 text{ eV}}{1.6 times 10^{-19} text{ J}}right) \
&= 4.14 text{ eV}
end{align*}
$$

Hence, this photon cannot eject electrons from gold

Step 2

2 of 3

b) We calculate the energy carried by this photon as follows:

$$
begin{align*}
E_2 &= hf_2 \
&= (6.626 times 10^{-34} text{ Js}) times (2.0 times 10^{15} text{ Hz}) \
&= 13.24 times 10^{-19} text{J} times left(dfrac{1 text{ eV}}{1.6 times 10^{-19} text{ J}}right) \
&= 8.28 text{ eV}
end{align*}
$$

Hence, this photon can eject electrons from gold

Result

3 of 3

a) Cannot eject electrons since $E_1 W_o$

Step 1

1 of 2

The energy of a photon has to be higher that the work function of the material in order for them to eject an electron. We will calculate the corresponding energies and compare them to $4.58text{ eV}$.

a) $E_1=hnu_1 = 6.62times10^{-34}text{ J s}times1.0times10^{15}text{ Hz} = 6.62times 10^{-19}text{ J}= frac{6.62times 10^{-19}}{1.6times 10^{-19}}text{ eV} = 4.14text{ eV}$ so this photon cannot eject an electron from gold surface.

b) $E_2=hnu_2 = 6.62times10^{-34}text{ J s}times2.0times10^{15}text{ Hz} = 13.24times 10^{-19}text{ J}= frac{13.24times 10^{-19}}{1.6times 10^{-19}}text{ eV} = 8.25text{ eV}$ so this photon can eject an electron from gold surface.

Result

2 of 2

Click here for the solution.

unlock

All Solutions

Page 861: Practice Problems

Haven't found what you were looking for?

Search for samples, answers to your questions and flashcards