Answer:
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
[tex]\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}[/tex]
c) Electrons required
[tex]\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}[/tex]
d) Kinetic energy of electrons
a) Energy of photon
[tex]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}[/tex]
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
a. The energy of molybdenum to eject electron has been [tex]\rm \bold{7.22\;\times\;10^{-19}}\;J[/tex].
b. The wavelength of the radiation has been 275 nm.
c. The number of electrons present in [tex]\rm 1\;\mu J[/tex] energy has been [tex]\rm \bold{1.38\;\times\;10^1^2}[/tex].
d. The kinetic energy of the emitted electrons as been [tex]\rm \bold{9.1\;\times\;10^-^1^9\;J}[/tex].
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:
[tex]E=h\nu[/tex]
Where, the value of constant, [tex]h=6.626\;\times\;10^{-34}\;\rm J.s[/tex]
The value of frequency has been given, [tex]\nu=\rm 1.09\;\times\;10^{15}\;s^{-1}[/tex]
Substituting the values for energy:
[tex]E=6.626\;\times\;10^{-34}\;\times\;1.09\;\times\;10^{15} \rm J\\\textit E=7.22\;\times\;10^{-19}\;J[/tex]
The energy of molybdenum to eject electron has been [tex]\rm \bold{7.22\;\times\;10^{-19}}\;J[/tex].
b. The wavelength ([tex]\lambda[/tex]) of the radiation has been given by:
[tex]\lambda=\dfrac{c}{\nu}[/tex]
Where, the speed of light, [tex]c=3\;\times\;\rm m/s[/tex]
The frequency has been given as, [tex]\nu=1.09\;\times\;10^{15}\;\rm s^-^1[/tex].
Substituting the values for wavelength:
[tex]\lambda=\dfrac{3\;\times\;10^8}{1.09\;\times\;10^{15}}\;\rm m\\ \lambda=2.75\;\times\;10^-^7\;m\\\lambda=275\;nm[/tex]
The wavelength of the radiation has been 275 nm.
c. The electrons ([tex]e^-[/tex]) burst out can be given as:
[tex]e^-=\dfrac{E}{E'}[/tex]
Where, the energy of radiations, [tex]E=1\;\times\;10^{-6}\;\rm J[/tex]
The energy of each electron has been calculated as, [tex]E'=7.22\;\times\;10^{-19}\;\rm J[/tex]
Substituting the values for number of electrons:
[tex]e^-=\dfrac{1\;\times\;10^-^6}{7.22\;\times\;10^-^1^9} \\e^-=1.38\;\times\;10^1^2[/tex]
The number of electrons present in [tex]\rm 1\;\mu J[/tex] energy has been [tex]\rm \bold{1.38\;\times\;10^1^2}[/tex].
d. The maximum kinetic energy (K.E.) of the radiation has been given as:
[tex]K.E.=\dfrac{hc}{\lambda} -E^'[/tex]
Where, the value of constant, [tex]h=6.626\;\times\;10^{-34}\;\rm J.s[/tex]
The speed of light, [tex]c=3\;\times\;\rm m/s[/tex]
The wavelength of the radiation, [tex]\lambda=122\;\times\;10^-^9\;\rm m[/tex]
Energy of each electron, [tex]E'=7.22\;\times\;10^{-19}\;\rm J[/tex]
Substituting the values, for kinetic energy:
[tex]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[/tex]
The kinetic energy of the emitted electrons as been [tex]\rm \bold{9.1\;\times\;10^-^1^9\;J}[/tex].
For more information about photoelectric effect, refer to the link:
https://brainly.com/question/9260704
