In a photoelectric experiment,the stopping po | vedclass

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(B) Einstein's photoelectric equation is given by: $e V = \frac{h c}{\lambda} - \phi$,which can be written as $\frac{h c}{\lambda} = \phi + e V$.For i

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$2 V_1+\frac{\phi}{e}$

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(B) Einstein's photoelectric equation is given by: $e V = \frac{h c}{\lambda} - \phi$,which can be written as $\frac{h c}{\lambda} = \phi + e V$.For incident wavelength $\lambda$,the stopping potential is $V_1$:$\frac{h c}{\lambda} = \phi + e V_1$  --- $(1)$For incident wavelength $\frac{\lambda}{2}$,the stopping potential is $V_2$:$\frac{h c}{\lambda / 2} = \phi + e V_2 \Rightarrow \frac{2 h c}{\lambda} = \phi + e V_2$  --- $(2)$Substitute the value of $\frac{h c}{\lambda}$ from equation $(1)$ into equation $(2)$:$2(\phi + e V_1) = \phi + e V_2$$2 \phi + 2 e V_1 = \phi + e V_2$$e V_2 = 2 e V_1 + \phi$Dividing by $e$,we get:$V_2 = 2 V_1 + \frac{\phi}{e}$

In a photoelectric experiment,the stopping potential was measured to be $V_1$ and $V_2$ volts with incident light of wavelength $\lambda$ and $\frac{\lambda}{2}$ respectively. The value of $V_2$ is: [where $\phi=$ work function,$e=$ electronic charge]

Similar Questions

In the photoelectric effect,the threshold wavelength of sodium is $5000 \ \mathring A$. Find its work function in $eV$. $(h = 6.6 \times 10^{-34} \ J \cdot s, c = 3 \times 10^8 \ m/s, 1 \ eV = 1.6 \times 10^{-19} \ J)$

Find the correct statement$(s)$ about the photoelectric effect.

The surface of a metal is first illuminated with a light of wavelength $300 \ nm$ and later illuminated by another light of wavelength $500 \ nm$. It is observed that the ratio of maximum velocities of photoelectrons in the two cases is $3$. The work function of the metal is close to: (in $eV$)

In the photoelectric effect,when photons of energy $h \nu$ fall on a photosensitive surface (work function $h \nu_0$),electrons are emitted from the metallic surface. It is possible to say that:

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