(A) According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E._{max})$ of emitted photoelectrons is given by:$K.E._{max} = h\nu

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Planck's constant,threshold frequency

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E._{max})$ of emitted photoelectrons is given by:$K.E._{max} = h\nu - \phi_0$Comparing this with the equation of a straight line $y = mx + c$,where $y = K.E._{max}$ and $x = \nu$:$1$. The slope $(m)$ of the graph is equal to Planck's constant $(h)$.$2$. The $X$-axis intercept occurs when $K.E._{max} = 0$,which gives $0 = h\nu_0 - \phi_0$,or $\nu_0 = \phi_0 / h$. This intercept represents the threshold frequency $(\nu_0)$.Therefore,the slope is Planck's constant and the $X$-axis intercept is the threshold frequency.

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

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The graph of maximum kinetic energy $(K.E._{max})$ against the frequency $(\nu)$ of incident light is as shown in the figure. The slope of the graph and the intercept on the $X$-axis respectively are:

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Planck's constant,threshold frequency
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Work function,maximum kinetic energy
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Planck's constant,work function

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Dual Nature of Radiation and matter

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Einstein's Photoelectric Equation and Energy Quantum of Radiation

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The work function of caesium is $2.14 \ eV$. Find
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If in a photoelectric experiment,the wavelength of incident radiation is reduced from $6000 \, \mathring{A}$ to $4000 \, \mathring{A}$,then:

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The graph of maximum kinetic energy $(K.E._{max})$ against the frequency $(\nu)$ of incident light is as shown in the figure. The slope of the graph and the intercept on the $X$-axis respectively are:

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The energy of the incident photon on a metal surface is $3W$ and then $5W$,where $W$ is the work function of that metal. The ratio of the maximum velocities of the emitted photoelectrons is:

When a photon of wavelength ${\lambda _1}$ is incident on a metal surface,the speed of the ejected fastest electron is $v_1 \ m/s$. If a photon of wavelength ${\lambda _2}$ is incident on the same metal,the speed of the ejected fastest electron is $v_2 \ m/s$. The value of $v_2^2 - v_1^2$ will be

The work function of caesium is $2.14 \ eV$. Find$(a)$ the threshold frequency for caesium,and$(b)$ the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of $0.60 \ V$.

If in a photoelectric experiment,the wavelength of incident radiation is reduced from $6000 \, \mathring{A}$ to $4000 \, \mathring{A}$,then:

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The work function of caesium is $2.14 \ eV$. Find
$(a)$ the threshold frequency for caesium,and
$(b)$ the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of $0.60 \ V$.