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son4ous [18]
3 years ago
11

Ultraviolet light of wavelength 3500 Angstrom falls on a potassium surface. The maximum energy of the emitted photoelectrons is

1.6 eV. Find the work function of potassium.
Physics
1 answer:
slega [8]3 years ago
8 0

Answer:

The work function of potassium is 1.94 eV.

Explanation:

Given that,

Wavelength \lambda= 3500\ \AA

Kinetic energy = 1.6 eV

We need to calculate the energy

Using formula of Energy

E=\dfrac{hc}{\lambda}

Put the value into the formula

E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{3500\times10^{-10}}

E=5.68\times10^{-19}\ J

E=5.68\times10^{-19}\times6.24\times10^{18}\ ev

E=3.54\ ev

We need to calculate the work function of potassium

Using formula of work function

E_{x}=\phi+k_{max}

Put the value into the formula

3.54=\phi+1.6

\phi=3.54-1.6

\phi=1.94\ ev

Hence, The work function of potassium is 1.94 eV.

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prohojiy [21]

Answer:

The capacite is C=5.32 uF using the equations of voltage and energy in capacitance  

Explanation:

The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J

Using:

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\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }

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ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }

C= 5.32x10^{-6} F

C= 5.32 uF because u is the symbol for micro that is equal to 10^{-6}

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