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Photoelectric work function of a metal is 1eV.Light of wavelength λ�� = 3000 Å falls on it. Thephoto electrons come out with velocity [1991]
- a)10 metres/sec
- b)102 metres/sec
- c)104 metres/sec
- d)106 metres/sec
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Photoelectric work function of a metal is 1eV.Light of wavelength &lam...
Solving we get, v ≌ 106 m/s
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Photoelectric work function of a metal is 1eV.Light of wavelength &lam...
Understanding the Photoelectric Effect
The photoelectric effect occurs when light shines on a metal surface, causing the emission of electrons. The energy of the incoming photons must exceed the work function of the metal for electrons to be emitted.
Given Data
- Work function (φ) = 1 eV
- Wavelength (λ) = 3000 Å = 3000 x 10^(-10) m = 3 x 10^(-7) m
Calculating Photon Energy
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) (Planck’s constant) = \( 6.626 \times 10^{-34} \, \text{Js} \)
- \( c \) (speed of light) = \( 3 \times 10^{8} \, \text{m/s} \)
Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^{8} \, \text{m/s})}{3 \times 10^{-7} \, \text{m}} \]
Calculating \( E \):
\[ E = \frac{(6.626 \times 3) \times 10^{-26}}{3} = 6.626 \times 10^{-19} \, \text{J} \]
Convert \( E \) into eV:
\[ E \approx \frac{6.626 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 4.14 \, \text{eV} \]
Finding Kinetic Energy of Electrons
Using the photoelectric equation:
\[ K.E. = E - φ \]
\[ K.E. = 4.14 \, \text{eV} - 1 \, \text{eV} = 3.14 \, \text{eV} \]
Converting \( K.E. \) to Joules:
\[ K.E. = 3.14 \times 1.6 \times 10^{-19} \, \text{J} \approx 5.024 \times 10^{-19} \, \text{J} \]
Calculating Velocity of Electrons
Using the kinetic energy formula:
\[ K.E. = \frac{1}{2} mv^2 \]
Assuming the mass of an electron, \( m = 9.11 \times 10^{-31} \, \text{kg} \):
\[ 5.024 \times 10^{-19} = \frac{1}{2} (9.11 \times 10^{-31}) v^2 \]
Solving for \( v \):
\[ v^2 = \frac{2 \times 5.024 \times 10^{-19}}{9.11 \times 10^{-31}} \approx 1.104 \times 10^{12} \]
\[ v \approx 10^6 \, \text{m/s} \]
Thus, the
The photoelectric effect occurs when light shines on a metal surface, causing the emission of electrons. The energy of the incoming photons must exceed the work function of the metal for electrons to be emitted.
Given Data
- Work function (φ) = 1 eV
- Wavelength (λ) = 3000 Å = 3000 x 10^(-10) m = 3 x 10^(-7) m
Calculating Photon Energy
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) (Planck’s constant) = \( 6.626 \times 10^{-34} \, \text{Js} \)
- \( c \) (speed of light) = \( 3 \times 10^{8} \, \text{m/s} \)
Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^{8} \, \text{m/s})}{3 \times 10^{-7} \, \text{m}} \]
Calculating \( E \):
\[ E = \frac{(6.626 \times 3) \times 10^{-26}}{3} = 6.626 \times 10^{-19} \, \text{J} \]
Convert \( E \) into eV:
\[ E \approx \frac{6.626 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 4.14 \, \text{eV} \]
Finding Kinetic Energy of Electrons
Using the photoelectric equation:
\[ K.E. = E - φ \]
\[ K.E. = 4.14 \, \text{eV} - 1 \, \text{eV} = 3.14 \, \text{eV} \]
Converting \( K.E. \) to Joules:
\[ K.E. = 3.14 \times 1.6 \times 10^{-19} \, \text{J} \approx 5.024 \times 10^{-19} \, \text{J} \]
Calculating Velocity of Electrons
Using the kinetic energy formula:
\[ K.E. = \frac{1}{2} mv^2 \]
Assuming the mass of an electron, \( m = 9.11 \times 10^{-31} \, \text{kg} \):
\[ 5.024 \times 10^{-19} = \frac{1}{2} (9.11 \times 10^{-31}) v^2 \]
Solving for \( v \):
\[ v^2 = \frac{2 \times 5.024 \times 10^{-19}}{9.11 \times 10^{-31}} \approx 1.104 \times 10^{12} \]
\[ v \approx 10^6 \, \text{m/s} \]
Thus, the
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Photoelectric work function of a metal is 1eV.Light of wavelength λ�� = 3000 Å falls on it. Thephoto electrons come out with velocity [1991]a)10 metres/secb)102 metres/secc)104 metres/secd)106 metres/secCorrect answer is option 'D'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Photoelectric work function of a metal is 1eV.Light of wavelength λ�� = 3000 Å falls on it. Thephoto electrons come out with velocity [1991]a)10 metres/secb)102 metres/secc)104 metres/secd)106 metres/secCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Photoelectric work function of a metal is 1eV.Light of wavelength λ�� = 3000 Å falls on it. Thephoto electrons come out with velocity [1991]a)10 metres/secb)102 metres/secc)104 metres/secd)106 metres/secCorrect answer is option 'D'. Can you explain this answer?.
Photoelectric work function of a metal is 1eV.Light of wavelength λ�� = 3000 Å falls on it. Thephoto electrons come out with velocity [1991]a)10 metres/secb)102 metres/secc)104 metres/secd)106 metres/secCorrect answer is option 'D'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Photoelectric work function of a metal is 1eV.Light of wavelength λ�� = 3000 Å falls on it. Thephoto electrons come out with velocity [1991]a)10 metres/secb)102 metres/secc)104 metres/secd)106 metres/secCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Photoelectric work function of a metal is 1eV.Light of wavelength λ�� = 3000 Å falls on it. Thephoto electrons come out with velocity [1991]a)10 metres/secb)102 metres/secc)104 metres/secd)106 metres/secCorrect answer is option 'D'. Can you explain this answer?.
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