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Step-by-Step Solution
Step 1: Write down the formula for the energy of an electron
The energy of an electron in the hydrogen atom according to the Bohr model is given by:
$E = -2.178 \times 10^{-18} \, J \Bigl(\frac{Z^2}{n^2}\Bigr)$
where $Z$ is the atomic number (for hydrogen, $Z = 1$) and $n$ is the principal quantum number.
Step 2: Identify the initial and final energy levels
The electron is excited from level $n_1 = 1$ to $n_2 = 2$. Hence, we need the energy difference
$\Delta E = E_2 - E_1$.
Step 3: Calculate the energy difference
Using the given formula, the energy for level $n$ is:
$E_n = -2.178 \times 10^{-18} \, J \Bigl(\frac{1^2}{n^2}\Bigr)$,
since $Z = 1$ for hydrogen.
Thus:
$E_1 = -2.178 \times 10^{-18} \, J \Bigl(\frac{1}{1^2}\Bigr) = -2.178 \times 10^{-18} \, J$
$E_2 = -2.178 \times 10^{-18} \, J \Bigl(\frac{1}{2^2}\Bigr)
= -2.178 \times 10^{-18} \, J \Bigl(\frac{1}{4}\Bigr)
= -0.5445 \times 10^{-18} \, J$
Therefore,
$\Delta E = E_2 - E_1
= \Bigl(-0.5445 \times 10^{-18}\Bigr) - \Bigl(-2.178 \times 10^{-18}\Bigr)
= 1.6335 \times 10^{-18} \, J$
Step 4: Relate the energy difference to the wavelength
The energy absorbed for this transition is given by the photon energy formula:
$\Delta E = \frac{hc}{\lambda}$
where $h$ is Planck's constant ($6.62 \times 10^{-34} \, \text{J} \cdot \text{s}$),
and $c$ is the speed of light ($3.0 \times 10^{8} \, \text{m/s}$).
Rewriting for $\lambda$:
$\lambda = \frac{hc}{\Delta E}$
Step 5: Substitute the numerical values and solve
$\lambda
= \frac{(6.62 \times 10^{-34} \, \text{J}\cdot \text{s}) \times (3.0 \times 10^8 \, \text{m/s})}
{1.6335 \times 10^{-18} \, \text{J}}$
$= 1.214 \times 10^{-7} \, \text{m}$
Final Answer
The wavelength of light required to excite the electron in hydrogen from $n=1$ to $n=2$
is $1.214 \times 10^{-7} \, \text{m}$.
