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1. An Alpha paricles moves in a circular path of radius 0.83cm in the presence of magnetic field of 0.25 wb/m². The wavelength associated with the particle will be a) 1A b)0.1 A c) 10A d)0.01 A?
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1. An Alpha paricles moves in a circular path of radius 0.83cm in the ...
Calculation of the Wavelength of Alpha Particle
Given:
Radius of circular path, r = 0.83 cm
Magnetic field, B = 0.25 Wb/m²
Formula:
The wavelength associated with the Alpha particle is given by λ = h/p, where h is Planck's constant and p is the momentum of the Alpha particle.
The momentum of the Alpha particle is given by p = qBr, where q is the charge of the Alpha particle, B is the magnetic field, and r is the radius of the circular path.
Therefore, λ = h/p = h/qBr
Solution:
Charge of Alpha particle, q = 2e (as it is a Helium nucleus)
where e is the charge of an electron.
Substituting the given values in the above formula, we get
λ = h/2eBr
Using the values of h = 6.626 x 10^-34 J.s, e = 1.6 x 10^-19 C, r = 0.83 cm = 0.0083 m and B = 0.25 Wb/m², we get
λ = (6.626 x 10^-34)/(2 x 1.6 x 10^-19 x 0.25 x 0.0083)
λ = 1.27 x 10^-10 m = 1.27 A
Answer:
The wavelength associated with the Alpha particle is 1.27 A. Therefore, the correct option is (a) 1A.
Given:
Radius of circular path, r = 0.83 cm
Magnetic field, B = 0.25 Wb/m²
Formula:
The wavelength associated with the Alpha particle is given by λ = h/p, where h is Planck's constant and p is the momentum of the Alpha particle.
The momentum of the Alpha particle is given by p = qBr, where q is the charge of the Alpha particle, B is the magnetic field, and r is the radius of the circular path.
Therefore, λ = h/p = h/qBr
Solution:
Charge of Alpha particle, q = 2e (as it is a Helium nucleus)
where e is the charge of an electron.
Substituting the given values in the above formula, we get
λ = h/2eBr
Using the values of h = 6.626 x 10^-34 J.s, e = 1.6 x 10^-19 C, r = 0.83 cm = 0.0083 m and B = 0.25 Wb/m², we get
λ = (6.626 x 10^-34)/(2 x 1.6 x 10^-19 x 0.25 x 0.0083)
λ = 1.27 x 10^-10 m = 1.27 A
Answer:
The wavelength associated with the Alpha particle is 1.27 A. Therefore, the correct option is (a) 1A.
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1. An Alpha paricles moves in a circular path of radius 0.83cm in the ...
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1. An Alpha paricles moves in a circular path of radius 0.83cm in the presence of magnetic field of 0.25 wb/m². The wavelength associated with the particle will be a) 1A b)0.1 A c) 10A d)0.01 A?
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1. An Alpha paricles moves in a circular path of radius 0.83cm in the presence of magnetic field of 0.25 wb/m². The wavelength associated with the particle will be a) 1A b)0.1 A c) 10A d)0.01 A? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 1. An Alpha paricles moves in a circular path of radius 0.83cm in the presence of magnetic field of 0.25 wb/m². The wavelength associated with the particle will be a) 1A b)0.1 A c) 10A d)0.01 A? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1. An Alpha paricles moves in a circular path of radius 0.83cm in the presence of magnetic field of 0.25 wb/m². The wavelength associated with the particle will be a) 1A b)0.1 A c) 10A d)0.01 A?.
1. An Alpha paricles moves in a circular path of radius 0.83cm in the presence of magnetic field of 0.25 wb/m². The wavelength associated with the particle will be a) 1A b)0.1 A c) 10A d)0.01 A? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 1. An Alpha paricles moves in a circular path of radius 0.83cm in the presence of magnetic field of 0.25 wb/m². The wavelength associated with the particle will be a) 1A b)0.1 A c) 10A d)0.01 A? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1. An Alpha paricles moves in a circular path of radius 0.83cm in the presence of magnetic field of 0.25 wb/m². The wavelength associated with the particle will be a) 1A b)0.1 A c) 10A d)0.01 A?.
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