(B) According to Einstein's photoelectric equation:$K_{\max} = h\nu - h\nu_0$Since $K_{\max} = eV_0$,where $V_0$ is the stopping potential and $e$ is

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Product of the slope of the line and the charge on the electron

(B) According to Einstein's photoelectric equation:$K_{\max} = h\nu - h\nu_0$Since $K_{\max} = eV_0$,where $V_0$ is the stopping potential and $e$ is the charge of an electron:$eV_0 = h\nu - h\nu_0$Dividing by $e$:$V_0 = (\frac{h}{e})\nu - (\frac{h\nu_0}{e})$Comparing this with the equation of a straight line $y = mx + c$,where $y = V_0$ and $x = \nu$:The slope $m = \frac{h}{e}$.Therefore,Planck's constant $h = m \times e$,which is the product of the slope of the line and the charge on the electron.

Class 12 Physics Dual Nature of Radiation and matter Einstein's Photoelectric Equation and Energy Quantum of Radiation

Medium

The stopping potential $V_0$ for photoelectric emission from a metal surface is plotted along the $Y-$ axis and the frequency $\nu$ of incident light along the $X-$ axis. $A$ straight line is obtained as shown. Planck's constant $h$ is given by:

A
Slope of the line
B
Product of the slope of the line and the charge on the electron
C
Product of the intercept along the $Y-$ axis and the mass of the electron
D
Product of the slope and the mass of the electron

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Dual Nature of Radiation and matter

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Einstein's Photoelectric Equation and Energy Quantum of Radiation

$A$ photosensitive surface has work function $\phi$. If a photon of energy $3 \phi$ falls on this surface, the electron comes out with a maximum velocity of $4 \times 10^6 \,m/s$. When the photon energy is increased to $7 \phi$, the maximum velocity of the photoelectron will be:

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Mercury violet light $(\lambda = 4558 \mathring{A})$ is falling on a photosensitive material $(\phi = 2.5 \text{ eV})$. The speed of the ejected electrons is,in $\text{m/s}$,about:

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The work functions of metals are given as $Na = 1.92 \, eV$,$K = 2.15 \, eV$,$Mo = 4.17 \, eV$,and $Ni = 5.0 \, eV$. Which of these metals will not show photoelectric emission for radiation of wavelength $3300 \, \mathring{A}$ from a $He-Cd$ laser?

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If light of wavelength $\lambda_1$ is allowed to fall on a metal,then the kinetic energy of the photoelectrons emitted is $E_1$. If the wavelength of light changes to $\lambda_2$,then the kinetic energy of the electrons changes to $E_2$. Then the work function of the metal is:

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Light of frequency $\nu$ is incident on a substance of threshold frequency $\nu_0$ (where $\nu_0 < \nu$). The maximum kinetic energy of the emitted photo-electron will be:

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The stopping potential $V_0$ for photoelectric emission from a metal surface is plotted along the $Y-$ axis and the frequency $\nu$ of incident light along the $X-$ axis. $A$ straight line is obtained as shown. Planck's constant $h$ is given by:

Similar Questions

$A$ photosensitive surface has work function $\phi$. If a photon of energy $3 \phi$ falls on this surface, the electron comes out with a maximum velocity of $4 \times 10^6 \,m/s$. When the photon energy is increased to $7 \phi$, the maximum velocity of the photoelectron will be:

Mercury violet light $(\lambda = 4558 \mathring{A})$ is falling on a photosensitive material $(\phi = 2.5 \text{ eV})$. The speed of the ejected electrons is,in $\text{m/s}$,about:

The work functions of metals are given as $Na = 1.92 \, eV$,$K = 2.15 \, eV$,$Mo = 4.17 \, eV$,and $Ni = 5.0 \, eV$. Which of these metals will not show photoelectric emission for radiation of wavelength $3300 \, \mathring{A}$ from a $He-Cd$ laser?

If light of wavelength $\lambda_1$ is allowed to fall on a metal,then the kinetic energy of the photoelectrons emitted is $E_1$. If the wavelength of light changes to $\lambda_2$,then the kinetic energy of the electrons changes to $E_2$. Then the work function of the metal is:

Light of frequency $\nu$ is incident on a substance of threshold frequency $\nu_0$ (where $\nu_0 < \nu$). The maximum kinetic energy of the emitted photo-electron will be:

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