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(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). (c) _(92)X.^(238)underset(-6beta)overset(-8alpha)(rarr) Y. Find out atomic number, mass number of Y and identify it. |
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Answer» `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)` |
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