Question

A spectral line is emitted when an atom undergoes transition between two levels with a difference in energy of 2.4eV. What is the wavelength of the line?

Question 10Select one:

287

n
m


377

Å


507

Å


518

n
m

Answer 1

To find the wavelength of the spectral line emitted when an atom undergoes a transition with an energy difference of 2.4 eV, we can use the formula that relates energy and wavelength:

\[
E = \frac{hc}{\lambda}
\]

where:
- \(E\) is the energy difference (in joules),
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J s}\)),
- \(c\) is the speed of light (\(3.00 \times 10^8 \, \text{m/s}\)),
- \(\lambda\) is the wavelength (in meters).

First, we need to convert the energy from electronvolts to joules. Since \(1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}\):

\[
E = 2.4 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = 3.843 \times 10^{-19} \, \text{J}
\]

Now, we can rearrange the formula to solve for \(\lambda\):

\[
\lambda = \frac{hc}{E}
\]

Substituting the values:

\[
\lambda = \frac{(6.626 \times 10^{-34} \, \text{J s}) \times (3.00 \times 10^8 \, \text{m/s})}{3.843 \times 10^{-19} \, \text{J}}
\]

Calculating this:

\[
\lambda = \frac{1.9878 \times 10^{-25}}{3.843 \times 10^{-19}} = 5.17 \times 10^{-7} \, \text{m} = 517 \, \text{nm}
\]

Converting meters to angstroms (1 nm = 10 Å):

\[
517 \, \text{nm} = 5170 \, \text{Å}
\]

Looking at the provided answers, assuming some rounding, the closest option is:

**518 nm** (which is equivalent to 5180 Å).

So the answer is:

**518 nm**