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If the stopping potential of a metal when illuminated with a radiation of wavelength 480 nm is 1.2 V, find (a) the work function of the metal, (b) the cutoff wavelength of the metal, and (c) the maximum energy of the ejected electrons

Respuesta :

Answer:

Part a)

[tex]W = 1.38 eV[/tex]

Part b)

[tex]\lambda = 901.22 nm[/tex]

Part c)

[tex]KE = 1.2 eV[/tex]

Explanation:

As we know by Einstein's equation of energy that

incident energy of photons = work function of metal + kinetic energy of electrons

here we know that incident energy of photons is given as

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

[tex]E = \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{480 \times 10^{-9}}[/tex]

now we have

[tex]E = 4.125 \times 10^{-19} J[/tex]

[tex]E = 2.58 eV[/tex]

kinetic energy of ejected electrons = qV

so we have

[tex]KE = e(1.2 V) = 1.2 eV[/tex]

Part a)

now we have

[tex]E = KE + W[/tex]

[tex]2.58 = 1.2 + W[/tex]

[tex]W = 1.38 eV[/tex]

Part b)

in order to find cut off wavelength we know that

[tex]W = \frac{hc}{\lambda}[/tex]

[tex]1.38 eV = \frac{1242 eV-nm}{\lambda}[/tex]

[tex]\lambda = 901.22 nm[/tex]

Part c)

Maximum energy of ejected electrons is the kinetic energy that we are getting

the kinetic energy of electrons will be obtained from stopping potential

so it is given as

[tex]KE = 1.2 eV[/tex]

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