The equation $$\lambda=\frac{1.227}{x} \mathrm{~nm}$$ can be used to find the de-Brogli wavelength of an electron. In this equation $$x$$ stands for :

Where

$$\mathrm{m}=$$ mass of electron

$$\mathrm{P}=$$ momentum of electron

$$\mathrm{K}=$$ Kinetic energy of electron

$$\mathrm{V}=$$ Accelerating potential in volts for electron

The activity of a radioactive material is $$6.4 \times 10^{-4}$$ curie. Its half life is 5 days. The activity will become $$5 \times 10^{-6}$$ curie after :

What is the half-life period of a radioactive material if its activity drops to $$1 / 16^{\text {th }}$$ of its initial value in 30 years?

A nucleus of mass $$M$$ at rest splits into two parts having masses $$\frac{M^{\prime}}{3}$$ and $${{2M'} \over 3}(M' < M)$$. The ratio of de Broglie wavelength of two parts will be :