Light of frequency v-5 v0 falls on a certain metal- then maximum velocity of electrons emitted is 8 - 106 m - s where v0 is threshold frequency of metal- If v-2 v0- then the maximum velocity of photoelectrons will beA- 4 - 106 m - sB- 3 - 106 m - sC- 1 - 106 m - sD- 2 - 106 m - s

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Standard XII

Chemistry

Planck's Quantum Hypothesis

Light of freq...

Question

Light of frequency $ν = 5 ν_{0}$ falls on a certain metal, then maximum velocity of electrons emitted is $8 \times 10^{6}$ m/s where $ν_{0}$ is threshold frequency of metal. If $ν = 2 ν_{0}$ , then the maximum velocity of photoelectrons will be

4 \times 10^{6}

m/s

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3 \times 10^{6}

m/s

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2 \times 10^{6}

m/s

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1 \times 10^{6}

m/s

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Solution

The correct option is A $4 \times 10^{6}$ m/s
Maximum kinetic energy depends upon frequency and work function.
$K_{1} = h (5 ν_{0} - ν_{0}) K_{2} = h (2 ν_{0} - ν_{0})$

Ratio of kinetic energy max is 4:1.
Hence speed is in ratio of 2:1.

New speed = $4 \times 10^{6}$ m/s

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Light of frequency ν=5ν0 falls on a certain metal, then maximum velocity of electrons emitted is 8×106 m/s where ν0 is threshold frequency of metal. If ν=2ν0, then the maximum velocity of photoelectrons will be

The correct option is A 4×106 m/sMaximum kinetic energy depends upon frequency and work function. K1=h(5ν0−ν0)K2=h(2ν0−ν0) Ratio of kinetic energy max is 4:1. Hence speed is in ratio of 2:1. New speed = 4×106 m/s

Light of frequency $ν = 5 ν_{0}$ falls on a certain metal, then maximum velocity of electrons emitted is $8 \times 10^{6}$ m/s where $ν_{0}$ is threshold frequency of metal. If $ν = 2 ν_{0}$ , then the maximum velocity of photoelectrons will be

The correct option is A $4 \times 10^{6}$ m/s
Maximum kinetic energy depends upon frequency and work function.
$K_{1} = h (5 ν_{0} - ν_{0}) K_{2} = h (2 ν_{0} - ν_{0})$

Ratio of kinetic energy max is 4:1.
Hence speed is in ratio of 2:1.

New speed = $4 \times 10^{6}$ m/s