The maximum kinetic energy of the photoelectrons is found to be 6-63 - 10-19 J- when the metal is irradiated with a radiation of frequency 2 - 1015 Hz The threshold frequency of the metal is about A- 1 X 1015 s-1B- 2 - 1015 s -1C- 3 - 1015 s-1D- 1-5 X 1015 s-1

The correct option is A K.E=h(v v_o) v_o=v K.E/h =2× 10^15 6.63× 10^ 19/6.63× 10^ 34 =1× 10^15 s^ 1 ...

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Standard XII

Chemistry

Planck's Quantum Hypothesis

The maximum k...

Question

The maximum kinetic energy of the photoelectrons is found to be $6.63 \times 10^{- 19} J$ , when the metal is irradiated with a radiation of frequency $2 \times 10^{15} H z$ The threshold frequency of the metal is about

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Solution

The correct option is A
$K . E = h (v - v_{o})$
$v_{o} = v - \frac{K . E}{h}$
$= 2 \times 10^{15} - \frac{6.63 \times 10^{- 19}}{6.63 \times 10^{- 34}}$
$= 1 \times 10^{15} s^{- 1}$

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The maximum kinetic energy of the photoelectrons is found to be 6.63×10−19 J, when the metal is irradiated with a radiation of frequency2×1015 Hz The threshold frequency of the metal is about

The correct option is A K.E=h(v−vo) vo=v−K.Eh =2×1015−6.63×10−196.63×10−34 =1×1015 s−1

The maximum kinetic energy of the photoelectrons is found to be $6.63 \times 10^{- 19} J$ , when the metal is irradiated with a radiation of frequency $2 \times 10^{15} H z$ The threshold frequency of the metal is about

The correct option is A
$K . E = h (v - v_{o})$
$v_{o} = v - \frac{K . E}{h}$
$= 2 \times 10^{15} - \frac{6.63 \times 10^{- 19}}{6.63 \times 10^{- 34}}$
$= 1 \times 10^{15} s^{- 1}$