According to classical physics, $10^{-15}\ m$ is distance of closest approach $(d_c)$ for fusion to occur between two protons. A more accurate and quantum approach says that ${d_c} = \frac{{{\lambda _p}}}{{\sqrt 2 }}$ where $'\lambda _p'$ is de-broglie's wavelength of proton when they were far apart. Using quantum approach, find equation of temperature at centre of star. [Given: $M_p$ is mass of proton, $k$ is boltzman constant]
$\frac{{{e^4}{M_p}}}{{24{\pi ^2}\varepsilon _0^2k{h^2}}}$
$\frac{{{e^4}{M_p}}}{{12{\pi ^2}\varepsilon _0^2k{h^2}}}$
$\frac{{{e^2}{M_p}}}{{24{\pi ^2}\varepsilon _0^2k{h^2}}}$
$\frac{{{e^4}{M_p}}}{{6{\pi ^2}\varepsilon _0^2k{h^2}}}$
The decay constant of a radio active substance is $0.173\, (years)^{-1}.$ Therefore :
A radio-isotope has a half- life of $5$ years. The fraction of the atoms of this material that would decay in $15$ years will be
If the radioactive decay constant of radium is $1.07 \times {10^{ - 4}}$ per year, then its half life period is approximately equal to .........$years$
At any instant the ratio of the amount of radioactive substances is $2 : 1$. If their half lives be respectively $12$ and $16$ hours, then after two days, what will be the ratio of the substances
Half life of radioactive element depends upon