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A particle A of mass 'm' and charge 'q' is accelerated by a potential difference of 50v another particle B of mass '4m' and charge 'q' is accelerated by a potential differnce of 2500V. The ratio of de - Broglie wavelength λA/λB is
Correct answer is '14.14'. Can you explain this answer?
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The de Broglie wavelength of a particle can be determined using the equation:
λ = h / p
where λ is the wavelength, h is the Planck's constant, and p is the momentum of the particle.
The momentum of a particle can be calculated using the equation:
p = mv
where m is the mass of the particle and v is its velocity.
For particle A:
The potential difference is 50V, so the velocity of particle A can be determined using the equation:
qV = 1/2 mv^2
50q = 1/2 mv^2
v^2 = 100q/m
v = √(100q/m)
The momentum of particle A can be calculated as:
p = mv
p = m * √(100q/m)
p = √(100mq)
For particle B:
The potential difference is 2500V, so the velocity of particle B can be determined using the same equation:
qV = 1/2 mv^2
2500q = 1/2 (4m) v^2
5000q = 4mv^2
v^2 = 5000q / 4m
v = √(5000q / 4m)
The momentum of particle B can be calculated as:
p = mv
p = (4m) * √(5000q / 4m)
p = √(5000mq)
Now, the ratio of the de Broglie wavelengths can be calculated as:
λ_A / λ_B = (h / p_A) / (h / p_B)
λ_A / λ_B = p_B / p_A
λ_A / λ_B = √(5000mq) / √(100mq)
λ_A / λ_B = √(50)
Therefore, the ratio of the de Broglie wavelengths of particle A to particle B is √(50) or approximately 7.071.
λ = h / p
where λ is the wavelength, h is the Planck's constant, and p is the momentum of the particle.
The momentum of a particle can be calculated using the equation:
p = mv
where m is the mass of the particle and v is its velocity.
For particle A:
The potential difference is 50V, so the velocity of particle A can be determined using the equation:
qV = 1/2 mv^2
50q = 1/2 mv^2
v^2 = 100q/m
v = √(100q/m)
The momentum of particle A can be calculated as:
p = mv
p = m * √(100q/m)
p = √(100mq)
For particle B:
The potential difference is 2500V, so the velocity of particle B can be determined using the same equation:
qV = 1/2 mv^2
2500q = 1/2 (4m) v^2
5000q = 4mv^2
v^2 = 5000q / 4m
v = √(5000q / 4m)
The momentum of particle B can be calculated as:
p = mv
p = (4m) * √(5000q / 4m)
p = √(5000mq)
Now, the ratio of the de Broglie wavelengths can be calculated as:
λ_A / λ_B = (h / p_A) / (h / p_B)
λ_A / λ_B = p_B / p_A
λ_A / λ_B = √(5000mq) / √(100mq)
λ_A / λ_B = √(50)
Therefore, the ratio of the de Broglie wavelengths of particle A to particle B is √(50) or approximately 7.071.
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A particle A of mass m and charge q is accelerated by a potential difference of 50v another particle B of mass 4m and charge q is accelerated by a potential differnce of 2500V. The ratio of de - Broglie wavelengthλA/λBisCorrect answer is '14.14'. Can you explain this answer?
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A particle A of mass m and charge q is accelerated by a potential difference of 50v another particle B of mass 4m and charge q is accelerated by a potential differnce of 2500V. The ratio of de - Broglie wavelengthλA/λBisCorrect answer is '14.14'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A particle A of mass m and charge q is accelerated by a potential difference of 50v another particle B of mass 4m and charge q is accelerated by a potential differnce of 2500V. The ratio of de - Broglie wavelengthλA/λBisCorrect answer is '14.14'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle A of mass m and charge q is accelerated by a potential difference of 50v another particle B of mass 4m and charge q is accelerated by a potential differnce of 2500V. The ratio of de - Broglie wavelengthλA/λBisCorrect answer is '14.14'. Can you explain this answer?.
A particle A of mass m and charge q is accelerated by a potential difference of 50v another particle B of mass 4m and charge q is accelerated by a potential differnce of 2500V. The ratio of de - Broglie wavelengthλA/λBisCorrect answer is '14.14'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A particle A of mass m and charge q is accelerated by a potential difference of 50v another particle B of mass 4m and charge q is accelerated by a potential differnce of 2500V. The ratio of de - Broglie wavelengthλA/λBisCorrect answer is '14.14'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle A of mass m and charge q is accelerated by a potential difference of 50v another particle B of mass 4m and charge q is accelerated by a potential differnce of 2500V. The ratio of de - Broglie wavelengthλA/λBisCorrect answer is '14.14'. Can you explain this answer?.
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