Home Page Textbooks Physics Page 862: Practice Problems

Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 862: Practice Problems

Exercise 11

Solution 1

Solution 2

Step 1

1 of 2

$textbf{Given:}$

$W_o = 2.28$ eV

$KE_{max} = 2.00$ eV

$textbf{Find: } f$

$textbf{Formula: } KE_{max} = hf – W_o$

Using this equation, we can obtain $f$ as follows:

$$
begin{align*}
KE_{max} &= hf – W_o \
f &= dfrac{KE_{max} + W_o}{h} \
&= dfrac{(2.00 text{ eV} + 2.28 text{ eV}) times left(dfrac{1.6 times 10^{-19} text{ eV}}{1 text{ eV}}right)}{6.626 times 10^{-34} text{ Js}} \
&= boxed{1.03 times 10^{15} text{ Hz}}
end{align*}
$$

Result

2 of 2

$f = 1.03 times 10^{15}$ Hz

Step 1

1 of 2

Using Einstein’s formula for the photoelectric effect we get

$$
hnu = W_0+E_k
$$
where $E_k$ is the kinetic energy of ejected electron. This yields

$$
nu = frac{W_0+E_k}{h} = frac{2.00text{ eV} + 2.28text{ eV}}{6.62times 10^{-34}text{ J s}} = frac{6.85times10^{-19}text{ J}}{6.62times10^{-34}text{ J s}} = 1.03times 10^{15}text{ Hz}.
$$

Result

2 of 2

Click here for the solution.

Exercise 12

Step 1

1 of 2

$textbf{Given:}$

$W_o = 4.58$ eV (as seen from Example 24.5)

$KE_{max} = 1.00$ eV

$textbf{Find: } f$

$textbf{Formula: } KE_{max} = hf – W_o$

Using this equation, we can obtain $f$ as follows:

$$
begin{align*}
KE_{max} &= hf – W_o \
f &= dfrac{KE_{max} + W_o}{h} \
&= dfrac{(1.00 text{ eV} + 4.58 text{ eV}) times left(dfrac{1.6 times 10^{-19} text{ eV}}{1 text{ eV}}right)}{6.626 times 10^{-34} text{ Js}} \
&= boxed{1.35 times 10^{15} text{ Hz}}
end{align*}
$$

Result

2 of 2

$$
f = 1.35 times 10^{15} text{ Hz}
$$

Exercise 13

Solution 1

Solution 2

Step 1

1 of 2

$textbf{Given:}$

$W_o = 2.5 text{ eV} times left(dfrac{1.6 times 10^{-19} text{ J}}{1 text{ eV}}right) = 4.0 times 10^{-19}$ J

$KE_{max} = 1.8 times 10^{-19} text{ J}$

$textbf{Find: } f$

$textbf{Formula: } KE_{max} = hf – W_o$

Using this equation, we can obtain $f$ as follows:

$$
begin{align*}
KE_{max} &= hf – W_o \
f &= dfrac{KE_{max} + W_o}{h} \
&= dfrac{(1.8 times 10^{-19} text{ J})+ (4.0 times 10^{-19} text{ J})} {6.626 times 10^{-34} text{ Js}} \
&= boxed{8.8 times 10^{14} text{ Hz}}
end{align*}
$$

Result

2 of 2

$$
f = 8.8 times 10^{14} text{ Hz}
$$

Step 1

1 of 2

Using Einstein’s formula for the photoelectric effect we get

$$
hnu = W_0+E_k
$$
where $E_k$ is the kinetic energy of ejected electron. This yields

$$
nu = frac{W_0+E_k}{h} = frac{2.5text{ eV} + 1.8times10^{-19}text{ J}}{6.62times 10^{-34}text{ J s}} = frac{4.0times10^{-19}text{ J} + 1.8times10^{-19}text{ J}}{6.62times10^{-34}text{ J s}} = frac{5.8times10^{-19}text{ J}}{6.62times10^{-34}text{ J s}} = 0.87times 10^{15}text{ Hz}.
$$

Result

2 of 2

Click here for the solution.

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