Answer
See the detailed answer below.
Work Step by Step
a) We can see in the figure below some possible modes of vibration of the wave of the particle confined to a one-dimensional box of width $L$.
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b) From the standing wave diagrams, in the figure below, we know that the wavelength is given by
$$\lambda =\dfrac{2L}{n}\tag 1$$
where $n=1,2,3,...$
And the momentum of this particle is given by
$$p=\dfrac{h}{\lambda}$$
Plugging $\lambda$ from (1);
$$p=\dfrac{hn}{2L}\tag 2$$
We know that kinetic energy is given by
$$KE=\frac{1}{2}mv^2=\frac{1}{2} v (mv)=\dfrac{pv}{2}$$
where $\lambda=\dfrac{h}{mv}$ and hence, $v=\dfrac{h}{m\lambda}$
Thus,
$$KE=\dfrac{p }{2}\;v=\dfrac{p }{2}\;\dfrac{h}{m\lambda}$$
Plugging from (1) and (2);
$$KE= \dfrac{hp}{2m\lambda}= \dfrac{h\dfrac{hn}{2L}}{2m\dfrac{2L}{n}}$$
$$\boxed{KE= \dfrac{h^2n^2 }{8mL^2}}$$
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c)
Inside the box, $E=KE+PE$, and the potential energy here is zero inside the box. So that $E=KE$.
Thus,
$$E_n= \dfrac{h^2n^2 }{8mL^2}$$
And for the ground-state energy where $n=1$;
$$E_1= \dfrac{h^2\times 1^2 }{8mL^2}$$
$$E_1= \dfrac{h^2 }{8m_eL^2e}\tag 3$$
where $e$ is the charge of the electron to find the energy in eV instead of J.
Plugging the known;
$$E_1= \dfrac{(6.626\times10^{-34})^2 }{8\times9.11\times10^{-31}\times (0.5\times10^{-10})^2\times 1.6\times 10^{-19}}$$
$$E_1=\color{red}{\bf 150.6}\;\rm eV$$
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d)
The ground state of the ball is given by the same formula we used for the electron above, but we will find it in joules so there is no need to divide by $e=1.6\times10^{-19}$
$$E_1= \dfrac{(6.626\times10^{-34})^2 }{8\times 140\times10^{-3 }\times (0.65)^2 }$$
$$E_1=\color{red}{\bf 9.278\times 10^{-67}}\;\rm eV$$
And its speed is given by
$$E_1=KE=\frac{1}{2}mv^2$$
Thus,
$$v=\sqrt{\dfrac{2E_1}{m}}$$
Plugging the known;
$$v=\sqrt{\dfrac{2\times 9.278\times 10^{-67}}{0.65}}$$
$$v=\color{red}{\bf 1.69\times 10^{-33}}\;\rm m/s$$
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e)
Solving (3) for $L$;
$$L^2 = \dfrac{h^2 }{8E_1m_ee} $$
$$L =\sqrt{ \dfrac{h^2 }{8E_1m_ee} }$$
$$L = \dfrac{h }{\sqrt{ 8E_1m_ee} }$$
Plugging the known;
$$L = \dfrac{6.626\times10^{-34}}{\sqrt{ 8 \times 22\times 1.6\times10^{-19}\times 9.11\times10^{-31}} }$$
$$L=\color{red}{\bf 1.31\times 10^{-10}}\;\rm m $$
