On a certain metal light of frequency ν=5ν0 then maximum velocity of electrons emitted is 8×106 ms-1, where ν0 is threshold frequency of the metal. If ν=2ν0 then the maximum velocity of of electron is x×106 ms-1. Find x.

Since Kmax=hν-hν0⇒ 12m8×1062=h5ν0-ν0=4hν0 ⇒ hν0=18m8×1062For the second case, we haveKE=12mv2=h2ν0-ν0=hν0=18m8×1062 ⇒ v=8×1062=4×106 ms-1

On a certain metal light of frequency ν=5ν0 then maximum velocity of electrons emitted is 8×106 ms-1, where ν0 is threshold frequency of the metal. If ν=2ν0 then the maximum velocity of of electron is x×106 ms-1. Find x.

Since Kmax=hν-hν0⇒ 12m8×1062=h5ν0-ν0=4hν0 ⇒ hν0=18m8×1062For the second case, we haveKE=12mv2=h2ν0-ν0=hν0=18m8×1062 ⇒ v=8×1062=4×106 ms-1

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On a certain metal light of frequency $ν = 5 ν_{0}$ then maximum velocity of electrons emitted is $8 \times 10^{6} {ms}^{- 1}$ , where $ν_{0}$ is threshold frequency of the metal. If $ν = 2 ν_{0}$ then the maximum velocity of of electron is $x \times 10^{6} {ms}^{- 1}$ . Find $x$ .

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Since $K_{max} = h ν - h ν_{0}$

$\Rightarrow \frac{1}{2} m {(8 \times 10^{6})}^{2} = h (5 ν_{0} - ν_{0}) = 4 h ν_{0} \Rightarrow h ν_{0} = \frac{1}{8} m {(8 \times 10^{6})}^{2}$

For the second case, we have

$K E = \frac{1}{2} m v^{2} = h (2 ν_{0} - ν_{0}) = h ν_{0} = \frac{1}{8} m {(8 \times 10^{6})}^{2} \Rightarrow v = \frac{8 \times 10^{6}}{2} = 4 \times 10^{6} {ms}^{- 1}$

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On a certain metal light of frequency ν=5ν0 then maximum velocity of electrons emitted is 8×106 ms-1, where ν0 is threshold frequency of the metal. If ν=2ν0 then the maximum velocity of of electron is x×106 ms-1. Find x.

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On a certain metal light of frequency $ν = 5 ν_{0}$ then maximum velocity of electrons emitted is $8 \times 10^{6} {ms}^{- 1}$ , where $ν_{0}$ is threshold frequency of the metal. If $ν = 2 ν_{0}$ then the maximum velocity of of electron is $x \times 10^{6} {ms}^{- 1}$ . Find $x$ .