(a) The work function of the surface is calculated as 22.177 x 10⁻²⁰ J = 1.384 eV and (b) The cutoff frequency for this surface is calculated as 3.347 x 10¹⁴ Hz.
Work function is the minimum thermodynamic work needed to remove electron from solid to a point in the vacuum immediately outside of the solid surface.
As the kinetic energy (KE) of the emitted photon is :
KE = 0.5mv²
and mass of electron, m = 9.1 X 10⁻³¹ kg
So, KE = 0.5 * 9.1 X 10⁻³¹ * (460000)²
= 9.628 X 10⁻²⁰ J
= 9.628 X 10⁻²⁰ x 6.242 X 10¹⁸
= 0.601 eV
The photon energy of the incoming radiation:
E = hf = hc/λ
As speed of light, c = 3 x 10⁸
Planck's constant, h = 6.626 × 10⁻³⁴ J.s
So now, E = (6.626 × 10⁻³⁴ *3 x 10⁸) /(625 X 10⁻⁹)
E = 31.805 X 10⁻²⁰ J
= 31.805 X 10⁻²⁰ x 6.242 X 10¹⁸
= 1.985 eV
(a) the work function of the surface is given by;
KE = hf - W
W = hf - KE; W is work function
Now, W = 31.805 X 10⁻²⁰ - 9.628 X 10⁻²⁰
= 22.177 x 10⁻²⁰ J
= 22.177 x 10⁻²⁰ x 6.242 X 10¹⁸
= 1.384 eV
(b) cutoff frequency is ;
W =hf
f = W/hf
= (22.177 x 10⁻²⁰ )/(6.626 × 10⁻³⁴ )
So, frequency = 3.347 x 10¹⁴ Hz
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