a) 2.3125 x 10-19 J
b) 3 x 10-19 J
c) 6.02 x 10-19 J
d) 6.62 x 10-34 J
When a radiation is passed on to the surface of a metal, some amount of energy is used in overcoming the attraction force on the electron and knocks it out from the atom. The remaining part is converted to kinetic energy.
Eradiation = Φ + KE
Φ = work function or ionization energy required to remove the electron from the atom.
KE = kinetic energy
Calculation of KE for one electron:
In the question, KE for one mole of electrons is given. However, in the options, the energy values are given for one electron. Hence we have to find the KE value for one electron.
From the data:
KE for 6.023 x 1023 electrons (one mole) = 6.023 x 104 J
KE for one electron = 6.023 x 104 / 6.023 x 1023 = 1 x 10-19 J
Calculation of Eradiation:
Eradiation = hc/λ = 6.626 x 10-34 J s x 3.0 x 108 m s-1/600 x 10-9 m = 3.313 x 10-19 J
Φ = Eradiation - KE = (3.313 x 10-19 J) - (1 x 10-19 J) = 2.313 x 10-19 J
The correct option is "a".
|Φ (in eV)||2.4||2.3||2.2||3.7||4.8||4.3||4.7||6.3||4.75|
(IIT JEE 2011)
Logic & solution:
The energy of radiation must be equal to or greater than the work functions of metals to show photoelectric effect.
We need to convert wavelength of radiation into energy expressed in eV units.
Eradiation = hc/λ = 6.626 x 10-34 J s x 3.0 x 108 m s-1/300 x 10-9 m = 6.626 x 10-19 J
Now convert this value into eV.
We know that:
1 J = 6.24 × 1018 eV
6.626 x 10-19 J = 6.626 x 10-19 x 6.24 × 1018 eV = 4.134 eV
Since the work functions of only Li, Na, K and Mg fall below 4.134 eV, only these metals can show photoelectric effect upon exposure of radiation of 300nm wavelength.
The number of metals that can show photoelectric effect = 4.
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