The velocity of the sodium electrons is [tex]3.5 \times 10^{5}[/tex] m/s
Given the following data:
To find the velocity of the sodium electrons, we would use the equation of photoelectric effect:
Mathematically, the photoelectric effect is given by the formula:
[tex]K.E = hf - BE_i[/tex]
Where:
First of all, we would determine the photon frequency of the light.
[tex]Photon\;frequency = \frac{speed}{wavelength} \\\\Photon\;frequency = \frac{3\times 10^8}{4.0 \times 10^{-7}}[/tex]
Photon frequency = [tex]7.5 \times 10^{14}[/tex] Hertz
Now, we can determine the kinetic energy from eqn 1:
[tex]K.E = 7.5 \times 10^{14}(6.626 \times 10^{-34}) - 4.4 \times 10^{-19}\\\\K.E = 4.97 \times 10^{-19} - 4.4 \times 10^{-19}[/tex]
K.E = [tex]5.70 \times 10^{-20}[/tex] Joules.
Also, the kinetic energy possessed by an electron is given by the formula:
[tex]K.E = \frac{1}{2} MV^2\\\\5.70 \times 10^{-20} = \frac{1}{2} \times 9.11 \times 10^{-31} \times V^2\\\\V = \sqrt{\frac{5.70 \times 10^{-20}}{4.56 \times 10^{-31}} }[/tex]
Velocity = [tex]3.5 \times 10^{5}[/tex] m/s
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