Q. 303.8( 5 Votes )

# Light of intensit

Answer :

Given:

Intensity of light, I = 10^{-5} Wm^{-2}

Surface area of photo-cell, A = 2 cm^{2} = 0.0002 m^{2}

Work function of sodium, Φ_{0} = 2eV = 2 × 1.6 × 10^{-19} = 3.2 × 10^{-19}J

By knowing the effective area of each sodium atom we can find the absorption of incident energy,

Effective area of each sodium atom, A_{Na} = 10^{-20} m^{2}

Number of layers of sodium, n = 5

Number of atoms absorbing radiation, N_{Na} = n × (A_{Na}/ A)

→ N_{Na} = 5 × (10^{-20} m^{2}/ 0.0002 m^{2})

→ N_{Na} = 10^{17}

So we conclude that 10^{17} atoms are effectively absorbing radiation.

Energy absorbed per atom, E = I/N_{Na}

E = 10^{-5}Js^{-1}/10^{17}

→ E = 2 × 10^{-26} Js^{-1}

Times required for photo electric emission, t is give by,

t = Φ_{0}/E

t = 3.2 × 10^{-19} J / 2 × 10^{-26} Js^{-1}

t = 1.6 × 10^{7} s

t = 0.5 years

Hence the time requires to initiate photoelectric emission is 0.5 years which is impractical hence, the wave model stands in disagreement with the experimental results.

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