Answer
See the detailed answer below.
Work Step by Step
a) We know that the uncertainty in energy is given by
$$\Delta E\Delta t\geq \hslash$$
Thus,
$$\Delta E\geq \dfrac{\hslash}{\Delta t}$$
Plugging the known;
$$\Delta E\geq \dfrac{1.055\times 10^{-34}}{10^{-8} }$$
$$\Delta E\geq \color{red}{\bf 1.055\times 10^{-34}} \;\rm J$$
$$\Delta E\geq\dfrac{ 1.055\times 10^{-34}}{1.6\times10^{-19}} $$
$$\Delta E\geq \color{red}{\bf 6.6\times10^{-8}} \;\rm eV$$
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b) We know, from Bohr hydrogen atom energies, that
$$E_n=\dfrac{-13.6Z^2}{n^2}$$
where $Z=1$;
$$E_n=\dfrac{-13.6 }{n^2}$$
Since the electron is back to $n=1$, the fraction transition energy is
$$\dfrac{\Delta E}{E_2-E_1}=\dfrac{ 6.6\times10^{-8}}{\frac{-13.6 }{2^2}-\frac{-13.6 }{1^2}}=\color{red}{\bf 6.47\times10^{-9}}$$
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c)
We know that the wavelength is given by
$$\Delta E_n=hf=\dfrac{hc}{\lambda}$$
Thus,
$$E=E_2-E_1=\dfrac{hc}{\lambda}$$
$$\lambda =\dfrac{hc}{(E_2-E_1 )e}$$
we multiplied the net energy by $e$ to convert the result for eV to J.
Plugging the known;
$$\lambda =\dfrac{6.626\times 10^{-34}\times3\times10^8}{\left(\frac{-13.6 }{2^2}-\frac{-13.6 }{1^2}\right)\times1.6\times10^{-19}}$$
$$\lambda= \color{red}{\bf 122}\;\rm nm$$
Now we need to find the width of this line in the hydrogen's spectrum.
$$\lambda+\Delta \lambda=\dfrac{hc}{(E_2-E_1 )+\Delta E }$$
$$\lambda+\Delta \lambda=\dfrac{hc}{E+\Delta E }=\dfrac{hc}{E\left(1+\frac{\Delta E}{E}\right) }$$
Thus,
$$\lambda+\Delta \lambda= \dfrac{hc}{E}\left(1+\frac{\Delta E}{E}\right)^{-1} $$
Applying the binomial expansion for $\left(1+\frac{\Delta E}{E}\right)^{-1}=\left(1-\frac{\Delta E}{E}\right) $
$$\lambda+\Delta \lambda= \dfrac{hc}{E}\left(1-\frac{\Delta E}{E}\right) $$
Noting that $ \dfrac{hc}{E}=\lambda$
$$\lambda+\Delta \lambda= \lambda\left(1-\frac{\Delta E}{E}\right) $$
$$\color{red}{\not}\lambda+\Delta \lambda= \color{red}{\not}\lambda -\lambda \frac{\Delta E}{E} $$
Therefore,
$$ |\Delta \lambda|= \lambda \frac{\Delta E}{E} $$
Plugging the known;
$$ |\Delta \lambda|= 122\frac{ 6.6\times10^{-8}}{ \frac{-13.6 }{2^2}-\frac{-13.6 }{1^2}} $$
$$ |\Delta \lambda|=\color{red}{\bf 789\times10^{-9}}\;\rm nm$$