# PHOTO ELECTRIC EFFECT

IIT JEE - NEET

AdiChemistry IIT JEE

## 1) Electrons with a kinetic energy of 6.023 x 10^{4} J/mol are evolved from the surface of a metal, when exposed to a radiation of wavelength of 600 nm (photoelectric effect). The minimum amount of energy required to remove an electron from the metal atom is :

#### (EAMCET
2009-E)

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

Logic:

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.

Therefore:

E_{radiation} = Φ + KE

Where:

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 10^{23} electrons (one mole) = 6.023 x 10^{4} J

KE for one electron = 6.023 x 10^{4} / 6.023 x 10^{23} = 1 x 10^{-19} J

**Calculation of E**_{radiation}:

E_{radiation} = hc/λ = 6.626 x 10^{-34} J s x 3.0 x 10^{8} m s^{-1}/600 x 10^{-9} m = 3.313 x 10^{-19} J

### Solution:

Φ = E_{radiation} - KE = (3.313 x 10^{-19} J) - (1 x 10^{-19} J) = 2.313 x 10^{-19} J

Conclusion:

The correct option is "a".

## Related questions

### 2) The work function (Φ) of some metals is listed below. The number of metals which will show photoelectric effect when light of 300nm wavelength falls on the metal is:

**Metal** |
**Li** |
**Na** |
**K** |
**Mg** |
**Cu** |
**Ag** |
**Fe** |
**Pt** |
**W** |

** 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.

E_{radiation} = hc/λ = 6.626 x 10^{-34} J s x 3.0 x 10^{8} 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 × 10^{18} eV

Therefore:

6.626 x 10^{-19} J = 6.626 x 10^{-19} x 6.24 × 10^{18} eV = 4.134 eV

Conclusion:

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.

AdiChemistry IIT JEE

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Author: Aditya vardhan Vutturi