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Given that work function of a metal is 5 eV and that photons of 9 eV are incident on it. the maximum velocity of emitted electrons is approximately______ x106 m/s (upto one decimal place)
Correct answer is between '1.1,1.3'. Can you explain this answer?
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Given that work function of a metal is 5 eV and that photons of 9 eV a...
Maximum kinetic energy o qf the electrons,
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Given that work function of a metal is 5 eV and that photons of 9 eV a...
The maximum velocity of emitted electrons can be determined using the concept of the photoelectric effect.
The photoelectric effect occurs when photons of sufficient energy are incident on a metal surface, causing the ejection of electrons. The energy required to remove an electron from the metal is known as the work function (Φ) of the metal.
In this case, the work function of the metal is given as 5 eV. This means that it requires a minimum energy of 5 eV to remove an electron from the metal surface.
The incident photons have an energy of 9 eV. To determine the maximum velocity of the emitted electrons, we need to calculate the excess energy of the photons after overcoming the work function.
Calculating the excess energy:
Excess energy = Energy of incident photons - Work function
Excess energy = 9 eV - 5 eV
Excess energy = 4 eV
The excess energy is the kinetic energy acquired by the emitted electron. To calculate the velocity of the electron, we can use the equation:
Kinetic energy = (1/2)mv^2
Where:
m = mass of the electron
v = velocity of the electron
Calculating the maximum velocity:
Since the mass of the electron (m) is a constant, we can rewrite the equation as:
Excess energy = (1/2)mv^2
Rearranging the equation to solve for velocity:
v^2 = (2 * Excess energy) / m
v = √((2 * Excess energy) / m)
Plugging in the values:
v = √((2 * 4 eV) / m)
The mass of an electron (m) is approximately 9.11 x 10^-31 kg.
v = √((2 * 4 eV) / (9.11 x 10^-31 kg))
Calculating the velocity:
v ≈ 1.24 x 10^6 m/s
Therefore, the maximum velocity of the emitted electrons is approximately 1.2 x 10^6 m/s.
Hence, the correct answer falls within the given range of 1.1 to 1.3 x 10^6 m/s.
The photoelectric effect occurs when photons of sufficient energy are incident on a metal surface, causing the ejection of electrons. The energy required to remove an electron from the metal is known as the work function (Φ) of the metal.
In this case, the work function of the metal is given as 5 eV. This means that it requires a minimum energy of 5 eV to remove an electron from the metal surface.
The incident photons have an energy of 9 eV. To determine the maximum velocity of the emitted electrons, we need to calculate the excess energy of the photons after overcoming the work function.
Calculating the excess energy:
Excess energy = Energy of incident photons - Work function
Excess energy = 9 eV - 5 eV
Excess energy = 4 eV
The excess energy is the kinetic energy acquired by the emitted electron. To calculate the velocity of the electron, we can use the equation:
Kinetic energy = (1/2)mv^2
Where:
m = mass of the electron
v = velocity of the electron
Calculating the maximum velocity:
Since the mass of the electron (m) is a constant, we can rewrite the equation as:
Excess energy = (1/2)mv^2
Rearranging the equation to solve for velocity:
v^2 = (2 * Excess energy) / m
v = √((2 * Excess energy) / m)
Plugging in the values:
v = √((2 * 4 eV) / m)
The mass of an electron (m) is approximately 9.11 x 10^-31 kg.
v = √((2 * 4 eV) / (9.11 x 10^-31 kg))
Calculating the velocity:
v ≈ 1.24 x 10^6 m/s
Therefore, the maximum velocity of the emitted electrons is approximately 1.2 x 10^6 m/s.
Hence, the correct answer falls within the given range of 1.1 to 1.3 x 10^6 m/s.
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Given that work function of a metal is 5 eV and that photons of 9 eV are incident on it. the maximum velocity of emitted electrons is approximately______ x106m/s (upto one decimal place)Correct answer is between '1.1,1.3'. Can you explain this answer?
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