(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi$,where $\phi$

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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.For $\lambda_1 = 800 \, nm$,$K_1 = \frac{1230}{800} - \phi = 1.5375 - \phi$ --- $(1)$For $\lambda_2 = 500 \, nm$,$K_2 = \frac{1230}{500} - \phi = 2.46 - \phi$ --- $(2)$Given that $K_2 = 2K_1$,we substitute the expressions:$2.46 - \phi = 2(1.5375 - \phi)$$2.46 - \phi = 3.075 - 2\phi$$2\phi - \phi = 3.075 - 2.46$$\phi = 0.615 \, eV$.

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

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$A$ metal is exposed to light of wavelength $800 \, nm$ and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of the photoelectrons doubles when light of wavelength $500 \, nm$ is used. The work function of the metal is $... \, eV$ (Take $hc = 1230 \, eV \cdot nm$).

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A
$1.537$
B
$2.46$
C
$0.615$
D
$1.23$

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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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$A$ metal is exposed to light of wavelength $800 \, nm$ and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of the photoelectrons doubles when light of wavelength $500 \, nm$ is used. The work function of the metal is $... \, eV$ (Take $hc = 1230 \, eV \cdot nm$).

Similar Questions

If two metals $A$ and $B$ are exposed to radiation of wavelength $350 \, nm$. The work functions of metals $A$ and $B$ are $4.8 \, eV$ and $2.2 \, eV$ respectively. Choose the correct option.

The work function of a material is $4.0 \ eV$. The maximum wavelength of light that can cause the emission of photoelectrons from the material is ............ $nm$.

According to Einstein's photoelectric equation,the graph of the maximum kinetic energy of emitted photoelectrons versus the frequency of incident radiation is a straight line. Its slope . . . . . .

At an incident radiation frequency of $v_1$,which is greater than the threshold frequency,the stopping potential for a certain metal is $V_1$. At frequency $2 v_1$,the stopping potential is $3 V_1$. If the stopping potential at frequency $4 v_1$ is $n V_1$,then $n$ is

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