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If an X-ray tube operates at the voltage of 10 kV, find the ratio of the de-Broglie wavelength of the incident electrons to the shortest wavelength of X-rays produced. The specific charge of an electron is 1.8 × 1011 C / kg.
- a)0.1
- b)0.4
- c)0.8
- d)1.0
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
If an X-ray tube operates at the voltage of 10 kV, find the ratio of t...
CONCEPT:
The wavelength of any charged particle due to its motion is called the de-Broglie wavelength.
When a charged particle is accelerated in a potential difference the energy gained by the particle is given by:
Energy (E) = e × V
Where V is the potential difference and q is a charge.
The de-Broglie wavelength of charge particle (λd) is given by:
Where E is energy, h is Planck constant, m is the mass of the charged particle
EXPLANATION:
De-Broglie wavelength when a charge q is accelerated by a potential difference of V volts is
For cut off wavele
ngth of X-rays, we have
From Eqs, i) and ii), we get
For electron q/m = 1.8 x 1011C/kg (given).
Substituting the values the desired ratio is
The wavelength of any charged particle due to its motion is called the de-Broglie wavelength.
When a charged particle is accelerated in a potential difference the energy gained by the particle is given by:
Energy (E) = e × V
Where V is the potential difference and q is a charge.
The de-Broglie wavelength of charge particle (λd) is given by:
Where E is energy, h is Planck constant, m is the mass of the charged particle
EXPLANATION:
De-Broglie wavelength when a charge q is accelerated by a potential difference of V volts is
For cut off wavele
ngth of X-rays, we haveFrom Eqs, i) and ii), we get
For electron q/m = 1.8 x 1011C/kg (given).
Substituting the values the desired ratio is
Most Upvoted Answer
If an X-ray tube operates at the voltage of 10 kV, find the ratio of t...
CONCEPT:
The wavelength of any charged particle due to its motion is called the de-Broglie wavelength.
When a charged particle is accelerated in a potential difference the energy gained by the particle is given by:
Energy (E) = e × V
Where V is the potential difference and q is a charge.
The de-Broglie wavelength of charge particle (λd) is given by:
Where E is energy, h is Planck constant, m is the mass of the charged particle
EXPLANATION:
De-Broglie wavelength when a charge q is accelerated by a potential difference of V volts is
For cut off wavelength of X-rays, we have
From Eqs, i) and ii), we get
For electron q/m = 1.8 x 1011C/kg (given).
Substituting the values the desired ratio is
The wavelength of any charged particle due to its motion is called the de-Broglie wavelength.
When a charged particle is accelerated in a potential difference the energy gained by the particle is given by:
Energy (E) = e × V
Where V is the potential difference and q is a charge.
The de-Broglie wavelength of charge particle (λd) is given by:
Where E is energy, h is Planck constant, m is the mass of the charged particle
EXPLANATION:
De-Broglie wavelength when a charge q is accelerated by a potential difference of V volts is
For cut off wavele
ngth of X-rays, we haveFrom Eqs, i) and ii), we get
For electron q/m = 1.8 x 1011C/kg (given).
Substituting the values the desired ratio is
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Community Answer
If an X-ray tube operates at the voltage of 10 kV, find the ratio of t...
To find the ratio of the de-Broglie wavelength of the incident electrons to the shortest wavelength of X-rays produced, we need to use the equation for the de-Broglie wavelength:
λ = h / (mv)
Where λ is the wavelength, h is Planck's constant (6.626 x 10^-34 J.s), m is the mass of the electron, and v is the velocity of the electron.
The specific charge of an electron is given by the equation:
q/m = specific charge
Where q is the charge of the electron (-1.6 x 10^-19 C) and m is the mass of the electron.
Given that the specific charge of an electron is 1.8 x 10^11 C/kg, we can solve for the mass of the electron:
(1.8 x 10^11 C/kg) = (-1.6 x 10^-19 C) / m
m = (-1.6 x 10^-19 C) / (1.8 x 10^11 C/kg)
m ≈ -8.89 x 10^-31 kg
Now, we can calculate the de-Broglie wavelength of the incident electrons:
λ = (6.626 x 10^-34 J.s) / ((-8.89 x 10^-31 kg) * v)
To find the shortest wavelength of X-rays produced, we can use the equation:
λ = c / f
Where λ is the wavelength, c is the speed of light (3 x 10^8 m/s), and f is the frequency of the X-rays.
The voltage of the X-ray tube is given as 10 kV, which is equal to 10,000 V. We can use this voltage to find the frequency of the X-rays:
f = (q * V) / h
f = (-1.6 x 10^-19 C * 10,000 V) / (6.626 x 10^-34 J.s)
Now we can calculate the shortest wavelength of the X-rays:
λ = (3 x 10^8 m/s) / ((-1.6 x 10^-19 C * 10,000 V) / (6.626 x 10^-34 J.s))
Finally, we can find the ratio of the de-Broglie wavelength to the shortest wavelength by dividing the two values:
Ratio = (de-Broglie wavelength) / (shortest wavelength)
λ = h / (mv)
Where λ is the wavelength, h is Planck's constant (6.626 x 10^-34 J.s), m is the mass of the electron, and v is the velocity of the electron.
The specific charge of an electron is given by the equation:
q/m = specific charge
Where q is the charge of the electron (-1.6 x 10^-19 C) and m is the mass of the electron.
Given that the specific charge of an electron is 1.8 x 10^11 C/kg, we can solve for the mass of the electron:
(1.8 x 10^11 C/kg) = (-1.6 x 10^-19 C) / m
m = (-1.6 x 10^-19 C) / (1.8 x 10^11 C/kg)
m ≈ -8.89 x 10^-31 kg
Now, we can calculate the de-Broglie wavelength of the incident electrons:
λ = (6.626 x 10^-34 J.s) / ((-8.89 x 10^-31 kg) * v)
To find the shortest wavelength of X-rays produced, we can use the equation:
λ = c / f
Where λ is the wavelength, c is the speed of light (3 x 10^8 m/s), and f is the frequency of the X-rays.
The voltage of the X-ray tube is given as 10 kV, which is equal to 10,000 V. We can use this voltage to find the frequency of the X-rays:
f = (q * V) / h
f = (-1.6 x 10^-19 C * 10,000 V) / (6.626 x 10^-34 J.s)
Now we can calculate the shortest wavelength of the X-rays:
λ = (3 x 10^8 m/s) / ((-1.6 x 10^-19 C * 10,000 V) / (6.626 x 10^-34 J.s))
Finally, we can find the ratio of the de-Broglie wavelength to the shortest wavelength by dividing the two values:
Ratio = (de-Broglie wavelength) / (shortest wavelength)
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