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Question: When electromagnetic radiation at wavelength 300nm falls on the surface of sodium, electrons are emi...

When electromagnetic radiation at wavelength 300nm falls on the surface of sodium, electrons are emitted with a kinetic energy of 1.68×105Jmol−11.68 \times {10^5}Jmo{l^{ - 1}}. What is the minimum energy needed to remove an electron from sodium? What is the maximum wavelength that will cause a photoelectron to be emitted?

Explanation

Solution

We have to first calculate the energy that is related with wavelength. We must focus on calculating the energy associated with one mole of photons and then multiply the answer by the Avogadro number.

Complete step by step answer:
We know that electromagnetic radiation is one form of energy that is present around us and is seen in several forms such as radio waves, x-rays, microwaves, gamma rays. Sunlight is an electromagnetic energy but visible light is a small part of the electromagnetic spectrum that has a wide range of electromagnetic wavelengths.
We can give the energy associated with 300nm photon as,
E=hcλE = \dfrac{{hc}}{\lambda }
Where,
h=Plank’s constant,
c=speed of light
λ\lambda =wavelength of light
We know the value of Planck's constant is 6.626×10−346.626 \times {10^{ - 34}} and value of wavelength of light is 3.0×108ms−13.0 \times {10^8}m{s^{ - 1}}
Let us substitute these values in the expression to calculate energy.
E=hcλE = \dfrac{{hc}}{\lambda }
Substituting the known values we get,
⇒E=(6.626×10−34Js)(3.0×108ms−1)300×10−19m\Rightarrow E = \dfrac{{\left( {6.626 \times {{10}^{ - 34}}Js} \right)\left( {3.0 \times {{10}^8}m{s^{ - 1}}} \right)}}{{300 \times {{10}^{ - 19}}m}}
⇒E=6.626×10−19J\Rightarrow E = 6.626 \times {10^{ - 19}}J
The energy associated with a 300nm photon is 6.626×10−19J6.626 \times {10^{ - 19}}J.
Let us now calculate the energy of one mole of photons using the Avogadro number.
E=(6.626×10−19J)(6.022×1023mol−1)E = \left( {6.626 \times {{10}^{ - 19}}J} \right)\left( {6.022 \times {{10}^{23}}mo{l^{ - 1}}} \right)
E=3.9×105Jmol−1E = 3.9 \times {10^5}Jmo{l^{ - 1}}
The energy of one mole of photons is calculated as 3.9×105Jmol−13.9 \times {10^5}Jmo{l^{ - 1}}.
Now let us calculate the minimum energy to remove a mole of electrons from sodium.
Minimum energy=(3.9×105Jmol−1)−(1.68×105Jmol−1)\left( {3.9 \times {{10}^5}Jmo{l^{ - 1}}} \right) - \left( {1.68 \times {{10}^5}Jmo{l^{ - 1}}} \right)
Minimum energy=2.31×105Jmol−12.31 \times {10^5}Jmo{l^{ - 1}}
The minimum energy to remove a mole of electrons from sodium is 2.31×105Jmol−12.31 \times {10^5}Jmo{l^{ - 1}}.
Let us calculate the minimum energy for one mole of electron using the minimum energy to remove a mole of electron from sodium and as,
Minimum energy=2.31×105Jmol−16.022×1023mol−1\dfrac{{2.31 \times {{10}^5}Jmo{l^{ - 1}}}}{{6.022 \times {{10}^{23}}mo{l^{ - 1}}}}
Minimum energy=3.84×10−19J3.84 \times {10^{ - 19}}J
The minimum energy for one mole of electrons is 3.84×10−19J3.84 \times {10^{ - 19}}J.
We know that,
E=hcλE = \dfrac{{hc}}{\lambda }
On rearranging the above equation, we get an expression to calculate wavelength.
λ=hcE\lambda = \dfrac{{hc}}{E}
Let us now substitute the values of planck's constant, wavelength of light and energy to get the wavelength
λ=hcE\lambda = \dfrac{{hc}}{E}
  ⟹  λ=(6.626×10−34)(3.0×108 ms−1)3.84×10−19J\implies \lambda = \dfrac{{\left( {6.626 \times {{10}^{ - 34}}} \right)\left( {3.0 \times {{10}^8}\,m{s^{ - 1}}} \right)}}{{3.84 \times {{10}^{ - 19}}J}}
∴λ=517nm\therefore \lambda = 517nm
The maximum wavelength that will cause photons to be emitted is 517nm517nm.
This wavelength comes under the region of green light. Hence, this could be maximum wavelength which causes the photon emission.

Note:
We must assume that the emission of electromagnetic waves is by electrically charged particles that undergo acceleration, and these waves interact with other charged particles exerting force on them. The effect of electromagnetic radiation on biological and chemical compounds depends on radiation power as well as its frequency.