In a photoelectric experiment for 4000 , math | vedclass

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K_{\max} = \frac{hc}{\lambda} - W_0$. Since the stopping po

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Greater than $2 \, V$

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K_{\max} = \frac{hc}{\lambda} - W_0$. Since the stopping potential $V_0$ is related to kinetic energy by $K_{\max} = eV_0$,we have $eV_0 = \frac{hc}{\lambda} - W_0$. Rearranging this,we get $V_0 = \frac{hc}{e\lambda} - \frac{W_0}{e}$. Here,$h$ is Planck's constant,$c$ is the speed of light,$e$ is the charge of an electron,$\lambda$ is the wavelength of incident light,and $W_0$ is the work function of the metal. As the wavelength $\lambda$ decreases from $4000 \, \mathring{A}$ to $3000 \, \mathring{A}$,the term $\frac{hc}{e\lambda}$ increases. Since $W_0$ is constant for a given metal,the stopping potential $V_0$ must increase. Therefore,the potential required to stop the ejection of electrons will be greater than $2 \, V$.

In a photoelectric experiment for $4000 \, \mathring{A}$ incident radiation,the potential difference to stop the ejection is $2 \, V$. If the incident light is changed to $3000 \, \mathring{A}$,then the potential required to stop the ejection of electrons will be

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