The work of Cesium is 2-27eV- The cut-off voltage which stops the emission of electrons from a cesium cathode irradiated with light of 600nm wavelength is0-5V-0-2V0-2VNone of the above-0-5V

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Toppr Admin · Accepted Answer

Energy of photon -hc-x3BB-2-066eVwhich is less than the work function of Cesium- Hence the concept of cut off voltage has no meaning here-

The work function of Cesium is $2.27 e V$ . The cut-off voltage which stops the emission of electrons from a cesium cathode irradiated with light of $600 n m$ wavelength is
$0.5 V$
$- 0.2 V$
$0.2 V$
None of the above
$- 0.5 V$

Energy of photon $= \frac{h c}{λ} = 2.066 e V$
which is less than the work function of Cesium. Hence the concept of cut off voltage has no meaning here.

The work function of Cesium is 2.27eV. The cut-off voltage which stops the emission of electrons from a cesium cathode irradiated with light of 600nm wavelength is0.5V−0.2V0.2VNone of the above−0.5V

Energy of photon =hcλ=2.066eVwhich is less than the work function of Cesium. Hence the concept of cut off voltage has no meaning here.

The work function of Cesium is $2.27 e V$ . The cut-off voltage which stops the emission of electrons from a cesium cathode irradiated with light of $600 n m$ wavelength is
$0.5 V$
$- 0.2 V$
$0.2 V$
None of the above
$- 0.5 V$

Energy of photon $= \frac{h c}{λ} = 2.066 e V$
which is less than the work function of Cesium. Hence the concept of cut off voltage has no meaning here.