The photoelectric threshold energy of certain metal is 3 eV. If light of wavelength 3000 $A^{o}$ is incident on the metal, then :

Electrons will be emitted

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Positrons will be emitted

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Protons will be emitted

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Electrons will not be emitted

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Solution

The correct option is B Electrons will be emitted
Energy of incident rays $= \frac{h c}{λ}$

$= \frac{1240}{300} (h c = 1240 e V - n m)$

$= 4.13 e V$
So, energy of incident rays is more than the threshold energy (3ev) so, electrons emission takes place.
So, the answer is option (A).

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The photoelectric threshold energy of certain metal is 3 eV. If light of wavelength 3000 Ao is incident on the metal, then :

The correct option is B Electrons will be emittedEnergy of incident rays =hcλ =1240300 (hc=1240eV−nm) =4.13eVSo, energy of incident rays is more than the threshold energy (3ev) so, electrons emission takes place.So, the answer is option (A).

The photoelectric threshold energy of certain metal is 3 eV. If light of wavelength 3000 $A^{o}$ is incident on the metal, then :

The correct option is B Electrons will be emitted
Energy of incident rays $= \frac{h c}{λ}$

$= \frac{1240}{300} (h c = 1240 e V - n m)$

$= 4.13 e V$
So, energy of incident rays is more than the threshold energy (3ev) so, electrons emission takes place.
So, the answer is option (A).