Answer:
To find the energy of an electron given its wavelength, we can use the de Broglie wavelength formula:
[tex]\lambda = \frac{h}{p}[/tex]
Where:
-[tex]\lambda [/tex] is the wavelength of the electron,
- [tex]h[/tex] is Planck's constant [tex]( 6.62607015 \times 10^{-34} \, \text{m}^2 kg/s[/tex]
-[tex]p[/tex]is the momentum of the electron.
The momentum of an electron can be calculated using its kinetic energy K and mass m:
[tex]p = \sqrt{2mK}[/tex]
Combining these equations, we get:
[tex]\lambda = \frac{h}{\sqrt{2mK}}[/tex]
Given [tex]\lambda = 5800, Å = 5800 \times 10^{-10} \,{m}[/tex],we can solve for [tex]K[/tex].
First, convert the wavelength to meters:
[tex]\[ 5800, {Å} = 5800 \times 10^{-10} m[/tex]
Now, rearrange the equation to solve for K
[tex]K = \frac{h^2}{2m\lambda^2}[/tex]
Plug in the values:
[tex][ K = \frac{(6.62607015 \times 10^{-34} , \text{m}^2 \cdot \text{kg} / \text{s})^2}{2 \times (9.10938356 \times 10^{-31} , \text{kg}) \times (5800 \times 10^{-10} , \text{m})^2} ][/tex]
Calculating this gives:
[tex]\[ K \approx 7.23 \times 10^{-25} \, \text{J} \][/tex]
Therefore, the correct answer is option [tex]**a. 7.23\times10^{-25}J**.[/tex]
