De Broglie Hypothesis

Given magnetic field, ,
Radius of the circular path,
We know radius of circular path in a uniform magnetic field,

and de -Broglie wavelength,
, For - particle
or



Hence, de-Broglie wavelength associated with -particle is .

The de-Broglie wavelength ( ) of a particle is related to its momentum ( ) by: , where is Planck's constant. So, . The kinetic energy (KE) of a non-relativistic particle is related to its momentum by: KE , where is the mass of the particle. Substituting : KE . Given values: Mass . de-Broglie wavelength . Planck's constant . Electric charge (this is the elementary charge, . It will be used for converting Joules to eV). Calculate KE in Joules first: KE . KE . KE . The terms cancel out: KE . KE . KE . Now, convert KE from Joules to electron-volts (eV). . So, KE (in eV) . KE (in eV) . KE (in eV) . . . So, KE . This value is approximately eV. This matches option (d).

1. We know that the de Broglie wavelength is related to the potential difference by the equation: Where:
is Planck's constant,
is the mass of the particle,
is the charge of the particle,
is the potential difference.
2. If the potential difference is increased by 21%, the new potential difference becomes .
3. Now, the de Broglie wavelength for the new potential difference is: 4. The ratio of the new wavelength to the original wavelength is: 5. Thus, the new wavelength is: 6. Therefore, the correct answer is , which corresponds to option (4).

The de Broglie wavelength of a particle is given by: where is Planck's constant, is the mass of the particle, and is its kinetic energy. Since the kinetic energy is the same for all particles, the particle with the smallest mass will have the largest wavelength. Therefore, the order of wavelengths is , where is the wavelength of the electron, the lightest particle.

For an electron in the first orbit of hydrogen, the de Broglie wavelength must be equal to the circumference of the orbit due to the standing wave condition.
Thus, the de Broglie wavelength is equal to the circumference of the orbit.

We know that the de Broglie wavelength is given by: where: - is Planck's constant, - is the mass of the particle, - is the velocity of the particle. Let be the velocity of the proton and be the velocity of particle A. From the problem, we know: Let be the mass of particle A. The ratio of the de Broglie wavelengths is: Substitute into the equation: Simplifying: Thus, the mass of particle A is .