Chemistry, 12.06.2020 05:57 coryowens44

Molybdenum metal requires a photon with a minimum frequency of 1.09x1015s-1before it can emit an electron via the photoelectric effect. a) What is the minimum energy needed to eject an electron?

b)What wavelength of radiation (in nm) will provide a photon of this energy?

c)How many electrons can be freed by a burst of radiation whose total energy is 1.00 μJ, assuming one photon causes one electron to be freed? (μ= micro = 10-6)

d) If molybdenum is irradiated with light of 122nm, what is the maximum kinetic energy of the emitted electrons?

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Answer from: Samzell

a. The energy of molybdenum to eject electron has been $\rm \bold{7.22\;\times\;10^{-19}}\;J$ .

b. The wavelength of the radiation has been 275 nm.

c. The number of electrons present in $\rm 1\;\mu J$ energy has been $\rm \bold{1.38\;\times\;10^1^2}$ .

d. The kinetic energy of the emitted electrons as been $\rm \bold{9.1\;\times\;10^-^1^9\;J}$ .

The metal emits the energy when it returns from the excited state to ground state.

a. The energy (E) to eject electron has been given by:

$E=h\nu$

Where, the value of constant, $h=6.626\;\times\;10^{-34}\;\rm J.s$

The value of frequency has been given, $\nu=\rm 1.09\;\times\;10^{15}\;s^{-1}$

Substituting the values for energy:

$E=6.626\;\times\;10^{-34}\;\times\;1.09\;\times\;10^{15} \rm J\\\textit E=7.22\;\times\;10^{-19}\;J$

The energy of molybdenum to eject electron has been $\rm \bold{7.22\;\times\;10^{-19}}\;J$ .

b. The wavelength ( $\lambda$ ) of the radiation has been given by:

$\lambda=\dfrac{c}{\nu}$

Where, the speed of light, $c=3\;\times\;\rm m/s$

The frequency has been given as, $\nu=1.09\;\times\;10^{15}\;\rm s^-^1$ .

Substituting the values for wavelength:

$\lambda=\dfrac{3\;\times\;10^8}{1.09\;\times\;10^{15}}\;\rm m\\ \lambda=2.75\;\times\;10^-^7\;m\\\lambda=275\;nm$

The wavelength of the radiation has been 275 nm.

c. The electrons ( e^- ) burst out can be given as:

$e^-=\dfrac{E}{E'}$

Where, the energy of radiations, $E=1\;\times\;10^{-6}\;\rm J$

The energy of each electron has been calculated as, $E'=7.22\;\times\;10^{-19}\;\rm J$

Substituting the values for number of electrons:

$e^-=\dfrac{1\;\times\;10^-^6}{7.22\;\times\;10^-^1^9} \\e^-=1.38\;\times\;10^1^2$

The number of electrons present in $\rm 1\;\mu J$ energy has been $\rm \bold{1.38\;\times\;10^1^2}$ .

d. The maximum kinetic energy (K.E.) of the radiation has been given as:

$K.E.=\dfrac{hc}{\lambda} -E^'$

Where, the value of constant, $h=6.626\;\times\;10^{-34}\;\rm J.s$

The speed of light, $c=3\;\times\;\rm m/s$

The wavelength of the radiation, $\lambda=122\;\times\;10^-^9\;\rm m$

Energy of each electron, $E'=7.22\;\times\;10^{-19}\;\rm J$

Substituting the values, for kinetic energy:

$K.E.=\dfrac{6.626\;\times\;10^{-34}\;\times\;3.0\;\times\;10^8}{122\;\times\;10^-^9}\;-\;7.22\;\times\;10^-^1^9\;\rm J \\\textit {K.E.}=9.1\;\times\;10^{-19}\;J$

The kinetic energy of the emitted electrons as been $\rm \bold{9.1\;\times\;10^-^1^9\;J}$ .

For more information about photoelectric effect, refer to the link:

link

Answer from: ninilizovtskt

a) 7.22 × 10⁻¹⁹ J; b) 275 nm; c) 1.38× 10¹² electrons; d) 9.1 × 10⁻¹⁹ J

Explanation:

a) Minimum energy to eject photon

E = hf = 6.626× 10⁻³⁴ J·s × 1.09 × 10¹⁵ s⁻¹ = 7.22 × 10⁻¹⁹ J

b) Wavelength required

fλ = c

$\lambda = \dfrac{c}{f } = \dfrac{2.998 \times 10^{8}\text{ m/s}}{1.09 \times 10^{15}\text{/s}} = 2.75 \times 10^{-7} \text{ m} = \textbf{275 nm}$

c) Electrons required

$\text{No. of electrons} = 1.00 \times 10^{-6}\text{ J} \times \dfrac{\text{1 electron}}{7.22 \times 10^{-19}\text{ J}} = 1.38 \times 10^{12}\text{ electrons}$

d) Kinetic energy of electrons

a) Energy of photon

$E = hf = \dfrac{\text{hc}}{\lambda} = \dfrac{6.626 \times 10^{-34} \text{ J$\cdot$ s}\times 2.998 \times 10^{8} \text{ m/s}}{122 \times 10^{-9}\text{ m}}= 1.63 \times 10^{-18} \text{ J}$

b) Maximum kinetic energy

The equation for the photoelectric effect is

hf = φ + KE, where

φ = the work function of the metal — the minimum energy needed to eject an electron

KE = hf - φ = 1.63× 10⁻¹⁸ J - 7.22× 10⁻¹⁹ J = 9.1 × 10⁻¹⁹ J

Answer from: Quest

Density because it’s holyyy

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Molybdenum metal requires a photon with a minimum frequency of 1.09x1015s-1before it can emit an ele...

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