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(B) The work function $\Phi_0$ of a metal is defined by the relation $\Phi_0 = \frac{hc}{\lambda_{th}}$,where $h$ is Planck's constant,$c$ is the spee

Vedclass · Accepted Answer

$2.28$

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(B) The work function $\Phi_0$ of a metal is defined by the relation $\Phi_0 = \frac{hc}{\lambda_{th}}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda_{th}$ is the threshold wavelength.Given $\lambda_{th} = 5420 \ \mathring A$.Using the shortcut formula $\Phi_0 (	ext{in } eV) = \frac{12400}{\lambda_{th} (	ext{in } \mathring A)}$.Substituting the value: $\Phi_0 = \frac{12400}{5420} \ eV$.$\Phi_0 \approx 2.28 \ eV$.

Light of wavelength $4000 \ \mathring A$ is incident on a sodium surface,where the threshold wavelength for photoemission is $5420 \ \mathring A$. The work function of sodium is ............ $eV$.

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