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Akimi4 [234]
3 years ago
10

When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a

potential difference of 1.43 V is needed to reduce the current to zero. What is the energy of a photon of this light in electron volts? energy of a photon: Find the work function of the irradiated metal in electron volts. work function:
Physics
1 answer:
sergij07 [2.7K]3 years ago
5 0

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 = \frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}

  = 4.59\times 10^{-19} \ J

or,

  = \frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}

  = 2.87 \ eV

(b)

As we know,

⇒ Vq=\frac{hc}{\lambda}-\Phi_0

By substituting the values, we get

⇒ 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

⇒                       \Phi_0=2.3\times 10^{-19} \ J

or,

⇒                            =\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}

⇒                            =1.4375 \ eV

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