Work, Energy and Power

Physics Part I

Physics

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Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in the *β*-decay of a neutron or a nucleus (Fig. 6.19).

[Note: The simple result of this exercise was one among the several arguments advanced by W. Pauli to predict the existence of a third particle in the decay products of*β*-decay. This particle is known as neutrino. We now know that it is a particle of intrinsic spin ½ (like *e*^{–}, *p *or *n*), but is neutral, and either massless or having an extremely small mass (compared to the mass of electron) and which interacts very weakly with matter. The correct decay process of neutron is:*n *→ *p + e*^{–}+ ν]

[Note: The simple result of this exercise was one among the several arguments advanced by W. Pauli to predict the existence of a third particle in the decay products of

The decay process of free neutron at rest is given as,

n

From Einstein’s mass-energy relation, we have

Energy of electron = Δ

where,

Δ

where, c = Speed of light

Δ

Hence, the given two-body decay is unable to explain the continuous energy distribution in the

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Is work a scalar or a vector quantity?

Work is dot product of two vectors. i.e. .

And dot product is a scalar quantity. Therefore, work is a scalar quantity.

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Define the unit joule.

Work done is said to be one joule if one newton of force displaces the body through a distance of one meter in the direction of applied force .

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