a piece of aluminum, with specific heat of 900 J/Kg K, loses 450 J of thermal en

Question

a piece of aluminum, with specific heat of 900 J/Kg K, loses 450 J of thermal energy, and its temperature changes from 25C to 20C. if the light with a frequency of 8.5x10^14 Hz shines on the piece of aluminum, what happen?

A. NO electrons are ejected, because the light's energy is too low.

B. more electrons are ejected than before the light shines on it.

C. electrons with more kinetic energy are ejected than before.

D. the same number of electrons, with the same kinetic energy, is ejected as before.

the right answer is C, BUT I DON'T KNOW WHY? I chose B.

CAN ANYONE SHOW ME HOW TO CALCULATE THE CUTOFF FREQUENCY?

Explanation / Answer

C. electrons with more kinetic energy are ejected than before.

use Einsteins photo euqtaion,

E = Wo + KE_max

here E --> energy of photon

Wo --> work function of the metal

KE_max - -> maximum kinteic energy of the ejected electrons.

we know,

E = h*f

as we increase the frequency Energy of photon increases.

so, maimum kinetic energy gained by elctrons also increases.

so, when we increase f, KE_max increases but no of elctrons ejected does not increase.

when light falls with least frequency, electrons are emitted with zero kinetic energy. This frequency is called cut-off frequency.

E = Wo + KE_max

h*f_cutoff = Wo + 0

f_cutoff = Wo/h (here h is plank's constant)

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