In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is :

Question

In photoelectric emission process from a metal of work 1-8 eV- the kinetic energy of energetic electrons is 0-5 eV- The corresponding stopping potential is -1-8 V1-3 V0-5 V2-3 V

Toppr Admin · Accepted Answer

Let V0 be the stopping potentialIn a photoelectric effect- the maximum kinetic energy of electron -xA0-Kmax-eV0so- V0-Kmaxe-0-5eVe-0-5V -xA0- -xA0-as eV-1 electronic charge times 1 Volt-

In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is :
1.8 V
1.3 V
0.5 V
2.3 V

Let $V_{0}$ be the stopping potential

In a photoelectric effect, the maximum kinetic energy of electron ,

$K_{m a x} = e V_{0}$

so, $V_{0} = \frac{K_{m a x}}{e} = \frac{0.5 e V}{e} = 0.5 V$ (as eV=1 electronic charge times 1 Volt)

In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is :1.8 V1.3 V0.5 V2.3 V

Let V0 be the stopping potentialIn a photoelectric effect, the maximum kinetic energy of electron , Kmax=eV0so, V0=Kmaxe=0.5eVe=0.5V (as eV=1 electronic charge times 1 Volt)

In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is :
1.8 V
1.3 V
0.5 V
2.3 V

Let $V_{0}$ be the stopping potential

In a photoelectric effect, the maximum kinetic energy of electron ,

$K_{m a x} = e V_{0}$

so, $V_{0} = \frac{K_{m a x}}{e} = \frac{0.5 e V}{e} = 0.5 V$ (as eV=1 electronic charge times 1 Volt)