The diagram shows the path of four $\alpha - $ particles of the same energy being scattered by the nucleus of an atom simultaneously. Which of these are/is not physically possible
The diagram shows the path of four $\alpha - $ particles of the same energy being scattered by the nucleus of an atom simultaneously. Which of these are/is not physically possible
- A
$3$ and $4$
- B
$2$ and $3$
- C
$1$ and $4$
- D
$4$ only
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Hydrogen $(H)$, deuterium $(D)$, singly ionized helium $(H{e^ + })$ and doubly ionized lithium $(Li)$ all have one electron around the nucleus. Consider $n =2$ to $n = 1 $ transition. The wavelengths of emitted radiations are ${\lambda _1},\;{\lambda _2},\;{\lambda _3}$ and ${\lambda _4}$ respectively. Then approximately
Hydrogen $(H)$, deuterium $(D)$, singly ionized helium $(H{e^ + })$ and doubly ionized lithium $(Li)$ all have one electron around the nucleus. Consider $n =2$ to $n = 1 $ transition. The wavelengths of emitted radiations are ${\lambda _1},\;{\lambda _2},\;{\lambda _3}$ and ${\lambda _4}$ respectively. Then approximately
What is head-on collision ? For that tell the impact parameter.
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When an $\alpha -$particle of mass $m$ moving with velocity $v$ bombards on a heavy nucleus of charge $Z_e$ , its distance of closest approach from the nucleus depends on $m$ as
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- [NEET 2016]
In an alpha particle scattering experiment distance of closest approach for the $\alpha$ particle is $4.5 \times 10^{-14} \mathrm{~m}$. If target nucleus has atomic number $80$ , then maximum velocity of $\alpha$-particle is . . . . .. $\times 10^5$ $\mathrm{m} / \mathrm{s}$ approximately.
$\left(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{SI}\right.$ unit, mass of $\alpha$ particle $=$ $\left.6.72 \times 10^{-27} \mathrm{~kg}\right)$
In an alpha particle scattering experiment distance of closest approach for the $\alpha$ particle is $4.5 \times 10^{-14} \mathrm{~m}$. If target nucleus has atomic number $80$ , then maximum velocity of $\alpha$-particle is . . . . .. $\times 10^5$ $\mathrm{m} / \mathrm{s}$ approximately.
$\left(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{SI}\right.$ unit, mass of $\alpha$ particle $=$ $\left.6.72 \times 10^{-27} \mathrm{~kg}\right)$
- [JEE MAIN 2024]
Give a powerful way to determine an upper limit to the size of the electron.
Give a powerful way to determine an upper limit to the size of the electron.