Home Page Textbooks Physics Page 867: Lesson Check

Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 867: Lesson Check

Exercise 26

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From De Broglie’s relation $lambda=h/p$ we see that the wavelength is inversely proportional to particle’s momentum so by increasing momentum we decrease the wavelength.

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Exercise 27

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He is correct because the baseball behaves only like a classical particle in a sense that it won’t show interference effects when it is fired to multiple slits and also does not show any other wave properties, unlike subatomic particles that show both particle and wave properties.

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Exercise 28

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This means that in some cases subatomic particles behave like classical particles i.e. they are localized, when they hit a screen they hit it at a single point etc. while in other cases they behave like classical waves i.e. they show interference patterns when go through multiple small slits.

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Exercise 29

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Because the energy of a single energy quantum of macroscopic objects is really small compared to macroscopic energies so we don’t even notice that the change is occurring in these small increments and appears as changing continuously.
Also macroscopic momenta are really large so from De Broglie’s relation $lambda = h/p$ we see that the wavelength is then really small and thus the wavelike effects are unnoticeable.

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Exercise 30

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It is less than that of the neutron.Since kinetic energy is related to the momentum in
$$
E=frac{p^2}{2m}
$$
which yields
$$
p=sqrt{2mE}
$$
we see that for fixed value of $E$ the greater the mass, the greater the momentum. Since by De Broglie’s relation wavelength is inversely proportional to the momentum the greater the momentum the smaller the wavelength so we finally conclude that since the neutron has greater mass that the electron, having the same kinetic energy it has smaller wavelength.

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Exercise 31

Solution 1

Solution 2

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$textbf{Given:}$

$lambda = 6.4 times 10^{-7}$ m

$textbf{Find:}$ speed of the proton, $v$

We use the de Broglie wave equation to calculate for $v$, such that:

$$
begin{align*}
lambda &= dfrac{h}{p} = dfrac{h}{mv} \
v &= dfrac{h}{m lambda} \
text{mass of the proton is} 1.67 times 10^{-24} text{ kg} \
&= dfrac{6.626 times 10^{-34} text{ Js}}{(1.67 times 10^{-24} text{ kg}) times (6.4 times 10^{-7} text{ m}} \
&= boxed{6.2 times 10^{-4} text{ m/s}}
end{align*}
$$

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$$
v = 6.2 times 10^{-4} text{ m/s}
$$

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From De Broglie’s relation we know that
$$
lambda = frac{h}{p}.
$$
Knowing that the mometum by definition is $p=mv$ we get

$$
lambda = frac{h}{mv}.
$$
Solving this for $v$ we obtain

$$
v=frac{h}{mlambda} = frac{6.62times10^{-34}text{ J s}}{1.67times 10^{-24}text{ kg}times 6.4times 10^{-7}text{ m}} =6.2times 10^{-4}text{ m/s}.
$$

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Exercise 32

Solution 1

Solution 2

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$textbf{Given:}$

$m = 6.6 times 10^{-27}$ kg

$v = 4.1 times 10^{4}$ m/s

$textbf{Find:}$ de Broglie wavelength, $lambda$

We use the de Broglie wave equation to calculate for $lambda$, such that:

$$
begin{align*}
lambda &= dfrac{h}{p} \
&= dfrac{h}{mv} \
&= dfrac{6.626 times 10^{-34} text{ Js}}{(6.6 times 10^{-27} text{ kg}) times (4.1 times 10^4 text{ m/s})} \
&= boxed{2.45 times 10^{-12} text{ m}}
end{align*}
$$

Result

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$$
lambda = 2.45 times 10^{-12} text{ m}
$$

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By De Broglie’s relation we have
$$
lambda = frac{h}{p}.
$$
Since by definition
$$
p=mv
$$
we get

$$
lambda = frac{h}{mv}.
$$
Note that the mass of helium atom is four times the mass of a proton so we have

$$
lambda =frac{6.62times10^{-34}text{ J s}}{6.6times10^{-27}text{ kg}times4.1times10^{4}text{ m/s}} = 2.45times10^{-12}text{ m}.
$$

Result

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Exercise 33

Solution 1

Solution 2

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$textbf{Given:}$

$lambda = 3.1 times 10^{-11}$ m

$textbf{Find:}$ Kinetic energy, $KE$

Energy can be calculated using the formula:

$$
begin{gather}
E = dfrac{p^2}{2m}
end{gather}
$$

We use the de Broglie wavelength equation to find a relationship between $lambda$ and $p$ as follows:

$$
begin{gather}
lambda = dfrac{h}{p} \
p = dfrac{h}{lambda}
end{gather}
$$

Combining Eq(1) and Eq(3) we can solve for $KE$:

$$
begin{align*}
KE &= dfrac{h^2}{2mlambda^2} \
&= dfrac{(6.626 times 10^{-34})^2}{2 times (1.67 times 10^{-24}) times (3.1 times 10^{-11})^2} \
&= 1.37 times 10^{-22} text{ J} \
text{converting this to eV:} \
&= 1.37 times 10^{-22} text{ J} times left(dfrac{1 text{ eV}}{1.6 times 10^{-19} text{ J}}right)\
&= boxed{8.5 times 10^{-4} text{ eV}}
end{align*}
$$

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$$
KE = 8.5 times 10^{-4} text{ eV}
$$

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First we will find the momentum from De Broglie’s relation $lambda=h/p$ which yields
$$
p=frac{h}{lambda}.
$$
Since the kinetic energy is related to the momentum by
$$
E=frac{p^2}{2m}
$$
putting in for $p$ we get

$$
E=frac{h^2}{2mlambda^2} = frac{(6.62times10^{-34}text{ J s})^2}{2times1.67times10^{-24}text{ kg }(3.1times10^{-11}text{ m})^2} = 1.37times10^{-22}text{ J}.
$$
This is in electron volts

$$
E=1.37times10^{-22}frac{1}{1.6times10^{-19}}text{ eV}=8.6times10^{ -4}text{ eV}.
$$

Result

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Exercise 34

Solution 1

Solution 2

Step 1

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$textbf{Given:}$

$p = 5.7 times 10^{-26} text{ kgm/s}$

$textbf{Formula:} lambda = dfrac{h}{p}$

a) Since the electron and proton both have equal momentum, their de Broglie wavelengths are also $boxed{text{equal}}$.

b) We calculate $lambda$ as follows:

$$
begin{align*}
lambda &= dfrac{6.626 times 10^{-34}}{5.7 times 10^{-26}} \
&= boxed{1.16 times 10^{-8} text{ m}}
end{align*}
$$

Result

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a) The de Broglie wavelength of the electron and proton are equal

b) $lambda = 1.16 times 10^{-8}$ m

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a) From De Broglie’s relation $lambda=h/p$ we see that the wavelength is inversely proportional to the momentum and is uniquely determined by it so they both have the same De Broglie’s wavelengths.

b) From the mentioned relation

$$
lambda = frac{h}{p} = frac{6.62times10^{-34}text{ J s}}{5.7times 10^{-26}text{ kg m/s}} =1.16times10^{-8}text{ m}.
$$

Result

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