The surface of sodium metal is irradiated wit | vedclass

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(D) According to the photoelectric equation: $E = \phi + K.E.$
First, convert the work function $\phi$ from eV to Joules: $\phi = 2.3 	ext{ eV} = 2.

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$5$

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(D) According to the photoelectric equation: $E = \phi + K.E.$
First, convert the work function $\phi$ from eV to Joules: $\phi = 2.3 	ext{ eV} = 2.3 	imes 1.602 	imes 10^{-19} 	ext{ J} \approx 3.6846 	imes 10^{-19} 	ext{ J}$.
Given kinetic energy $K.E. = 2.8 	imes 10^{-20} 	ext{ J} = 0.28 	imes 10^{-19} 	ext{ J}$.
Total energy of the incident radiation $E = 3.6846 	imes 10^{-19} + 0.28 	imes 10^{-19} = 3.9646 	imes 10^{-19} 	ext{ J}$.
Using the relation $E = \frac{hc}{\lambda}$, where $h = 6.626 	imes 10^{-34} 	ext{ J}\cdot	ext{s}$ and $c = 3 	imes 10^8 	ext{ m/s}$:
$\lambda = \frac{hc}{E} = \frac{6.626 	imes 10^{-34} 	imes 3 	imes 10^8}{3.9646 	imes 10^{-19}} \approx 5.01 	imes 10^{-7} 	ext{ m} = 501 	ext{ nm}$.
Rounding to the nearest integer as per the format, $x = 5$, so $x = 5 	imes 10^2$ nm.

The surface of sodium metal is irradiated with radiation of wavelength $x$ nm. The kinetic energy of ejected electrons is $2.8 \times 10^{-20}$ $J$. The work function of sodium is $2.3$ eV. The value of $x$ is . . . . . . $\times 10^2$ nm.

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