(a) The wave function of an electron in `2s` orbital in hydrogen atom is given below:
`psi_(2s)=1/(4(2pi)^(1//2))(z/a_(0))^(3//2)(2-r/a_(0))exp(-r//2a_(0))`
where `a_(0)` is the radius. This wave function has a radial node at `r=r_(0)`. Express `r_(0)` in terms of `a_(0)`.
(b) Calculate the wavelength of a ball of mass `100 g` moving with a velocity of `100 ms^(-1)`.

Answer

(a) `R=2A_(0)`, (b) `lambda=6.626xx10^(-25) Å` (c) `_(82)Y^(206);`(Atomic no. 82, Mass no. 206)

Explanation

`Psi_(2s)^(2)`=Probability of finding electron with in `2s` orbital
`Psi_("at node")^(2)=0` (probability of finding an electron is zero at node)
for node at `r=r_(0), Psi^(2)=0`
So, `Psi^(2)=0=(1)/(4sqrt(2pi))[1/a_(0)]^(3), [2-r_(0)/a_(0)]xxe^(r_(0)//2a_(0))rArr [2-r_(0)/a_(0)]=0` or `2=r/a_(0)`
`rArr r=2a_(0)`
(b) The wavelength can be calculate with the help of de-Broglie's formula i.e.,
`lambda=h/(mv)=(6.626xx10^(-34))/(100xx100xx10^(-3))=(6.626xx10^(-34))/(10,000xx10^(-3))=6.626xx10^(-35) m` or `6.626xx10^(-25) Å`
(c) (i) The atomic mass of element reduces by `4` and atomic number by `2` on emission of an `alpha`-particle.
(ii) the atomic mass of an element remains unchanged and atomic increases by `1` on emission of a `beta`-particle.
Thus change in atomic mass on emission of `8alpha`-particles will be `8xx4=32`
New atomic mass = old atomic mass `-32=238-32=206`
similarly change in atomic number on emission of `8alpha`-particle will be: `8xx2=16`
i.e., New atomic number = old atomic `-16=92-16=76`
On emission of `6beta`-particles the atomic mass remains unchanged thus, atomic mass of the new element will be `206`.
The atomic number increases by `6` unit thus new atomic number will be `76+6=82`
Thus, the equation looks like: `._(92)X.^(238) underset(-6beta)overset(-8alpha)(rarr)._(82)Y^(206)`