Answer:
The right solution is:
(a) 2.87 eV
(b) 1.4375 eV
Explanation:
Given:
Wavelength,
= 433 nm
Potential difference,
= 1.43 V
Now,
(a)
The energy of photon will be:
E = [tex]\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}[/tex]
= [tex]4.59\times 10^{-19} \ J[/tex]
or,
= [tex]\frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}[/tex]
= [tex]2.87 \ eV[/tex]
(b)
As we know,
⇒ [tex]Vq=\frac{hc}{\lambda}-\Phi_0[/tex]
By substituting the values, we get
⇒ [tex]1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0[/tex]
⇒ [tex]\Phi_0=2.3\times 10^{-19} \ J[/tex]
or,
⇒ [tex]=\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}[/tex]
⇒ [tex]=1.4375 \ eV[/tex]
