The work function of a metallic surface is 5. | vedclass

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Question

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the emitted photoelectrons is given by $K_{max} = E - \phi$

Vedclass · Accepted Answer

$1.2$

Vedclass · Answer

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the emitted photoelectrons is given by $K_{max} = E - \phi$,where $E$ is the energy of the incident photon and $\phi$ is the work function of the metal surface.The energy of the incident photon $E$ is calculated as $E = \frac{12400}{\lambda(	ext{in } \mathring{A})} \ eV$.Given $\lambda = 2000\ \mathring{A}$,we have $E = \frac{12400}{2000} = 6.2\ eV$.The work function $\phi = 5.01\ eV$.Therefore,$K_{max} = 6.2\ eV - 5.01\ eV = 1.19\ eV \approx 1.2\ eV$.The stopping potential $V_S$ is related to the maximum kinetic energy by $K_{max} = eV_S$.Thus,$eV_S = 1.2\ eV$,which implies $V_S = 1.2\ V$.

The work function of a metallic surface is $5.01\ eV$. Photoelectrons are emitted when light of wavelength $2000\ \mathring{A}$ falls on it. The minimum potential difference required to stop the fastest photoelectrons is ................. $V$.

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