
In $L{i^{ + + }}$, electrons in the first Bohr orbit are excited to a level by radiation of wavelength $\lambda $. When the ion gets deexcited to the ground state in all possible ways (including intermediate emissions), a total of 6 spectral lines are observed. What is the value of $\lambda $?
(Given: $h = 6.63 \times {10^{34}}Js$; $c = 3 \times {10^8}m/s$)
(a) 9.4 nm
(b) 12.3 nm
(c) 10.8 nm
(d) 11.4 nm
Answer
595.8k+ views
Hint:
The spectral lines observed can be used to find the transition level of the atom. As transition energies are fixed for energy levels, this can be used to determine $\lambda $.
Formula used:
Number of spectral lines is given by:
No. of spectral lines$ = n\dfrac{{(n - 1)}}{2}$ ……(1)
where n is the transition level.
Change in energy for transition between energy levels ${n_1}$ and ${n_2}$ is given by:
$\Delta E = 13.6 \times {Z^2}(\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}})$ ……(2)
where Z is the atomic number of the atom.
Energy is related to wavelength by:
$\Delta E = hc/\lambda $ ……(3)
Complete step-by-step answer:
Given:
1. Atom is $L{i^{ + + }}$ with $Z = 3$.
2. No. of spectral lines = 6.
To find: The wavelength $\lambda $ which excited the atom.
Step 1 of 3:
Find the energy level transition from no. of spectral lines using eq (1):
$\begin{gathered}
6 = n\dfrac{{(n - 1)}}{2} \\
{n^2} - n - 12 = 0 \\
(n - 4)(n + 3) = 0 \\
n = 4, - 3 \\
\end{gathered} $
As n cannot be negative, n is 4. This means the transition occurred from ${n_1} = 1$ to \[{n_2} = 4\].
Step 2 of 3:
Find $\Delta E$ for transition from ${n_1} = 1$ to \[{n_2} = 4\] using eq (2):
$\begin{gathered}
\Delta E = 13.6 \times {3^2}(\dfrac{1}{{{1^2}}} - \dfrac{1}{{{4^2}}}) \\
\Delta E = 13.6 \times 9 \times \dfrac{{15}}{{16}} \\
\Delta E = 114.75eV \\
\end{gathered} $
Step 3 of 3:
Find $\lambda $ by putting $hc = 1240eV.nm$ in eq (3):
$\begin{gathered}
114.75eV = 1240eV.nm/\lambda \\
\lambda = 10.8nm \\
\end{gathered} $
Correct Answer:
The wavelength $\lambda $ which excited the atom: (c) 10.8 nm
Note:In questions like these first find the transition level. Then find the corresponding energy for this transition. From this energy, we can find the wavelength.
The spectral lines observed can be used to find the transition level of the atom. As transition energies are fixed for energy levels, this can be used to determine $\lambda $.
Formula used:
Number of spectral lines is given by:
No. of spectral lines$ = n\dfrac{{(n - 1)}}{2}$ ……(1)
where n is the transition level.
Change in energy for transition between energy levels ${n_1}$ and ${n_2}$ is given by:
$\Delta E = 13.6 \times {Z^2}(\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}})$ ……(2)
where Z is the atomic number of the atom.
Energy is related to wavelength by:
$\Delta E = hc/\lambda $ ……(3)
Complete step-by-step answer:
Given:
1. Atom is $L{i^{ + + }}$ with $Z = 3$.
2. No. of spectral lines = 6.
To find: The wavelength $\lambda $ which excited the atom.
Step 1 of 3:
Find the energy level transition from no. of spectral lines using eq (1):
$\begin{gathered}
6 = n\dfrac{{(n - 1)}}{2} \\
{n^2} - n - 12 = 0 \\
(n - 4)(n + 3) = 0 \\
n = 4, - 3 \\
\end{gathered} $
As n cannot be negative, n is 4. This means the transition occurred from ${n_1} = 1$ to \[{n_2} = 4\].
Step 2 of 3:
Find $\Delta E$ for transition from ${n_1} = 1$ to \[{n_2} = 4\] using eq (2):
$\begin{gathered}
\Delta E = 13.6 \times {3^2}(\dfrac{1}{{{1^2}}} - \dfrac{1}{{{4^2}}}) \\
\Delta E = 13.6 \times 9 \times \dfrac{{15}}{{16}} \\
\Delta E = 114.75eV \\
\end{gathered} $
Step 3 of 3:
Find $\lambda $ by putting $hc = 1240eV.nm$ in eq (3):
$\begin{gathered}
114.75eV = 1240eV.nm/\lambda \\
\lambda = 10.8nm \\
\end{gathered} $
Correct Answer:
The wavelength $\lambda $ which excited the atom: (c) 10.8 nm
Note:In questions like these first find the transition level. Then find the corresponding energy for this transition. From this energy, we can find the wavelength.
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