why does n/p ratio inrease in the case of ₉₂U²³⁸?it emits alpha particle.but why?

This is such a great question!!! But be warned great questions rarely have simple answers.

Your wondering why Uranium is unstable and why it emits alpha particles. But what is stabillity? The half life of U238 is ~10⁹ years. Depending on who you talk to thats a very stable isotope. So, where do we draw the line on what is called "stable" or what isn't? When Uranium undergoes alpha decays it will emit helium nucleii and eventually become Lead 206. Lead 206 has a higher binding energy per nucleon. Meaning that the protons and neutrons are held more tightly in a  lead atom than an uranium atom. That being said, Uranium has a higher binding energy per nucleon than Helium! Helium is considered to be highly stable, why the contradiction? What would be a better way to measure nuclear stabillity? I'll leave the discussion open-ended so others can expand on it.

the Uranium is under the stabilitybelt.

but (238-92)/92=1.586. So,how can it be below the stability belt?

As a side note (and one the Mitch has asked me to add), even though the n/p ratio has increased the stability of the new nucleus has decreased.

why in the 1st place shud the ratio increase?wouldn't the daughter nuclide be farther away from the stability zone

but (238-92)/92=1.586. So,how can it be below the stability belt? ???why in the 1st place shud the ratio increase?wouldn't the daughter nuclide be farther away from the stability zone

Check Grejak's math - it is as simple as counting using fingers

Put aside all stability/unstability, binding energy per nucleon and similar problems. Look just for the quotient:

n/p

n is - at the moment of start - 146, p is 92

when the alpha particle is emitted you substract 2 from both numerator and denominator. As the denominator is smaller such substraction makes quotient larger. This is very simple math, nothing fancy, and nothing to do with the radiochemistry.

Now, if you want to know why alpha particle is emitted, you have to look for stabilities and so on, but that's completely different story an it requires knowledge. Mitch is much better here then me. I am using only common reasoning

yes but shud it be that way?it shud have tried to bring down the ratio.

I think your question it more specific than you want it to be.  You asked why alpha decay in specific increases the n/p ratio in a nucleus.  That is simple number manipulation and has been sufficiently answered multiple times.  No matter how many times you ask the question, alpha decay will always increase the n/p ratio in ²³⁸U.  Not due to the stability of the nucleus or any other factors, simply due to the definition of what alpha decay is.  That cannot be changed.

So what can be changed?  Well, the decay mode does not always have to be alpha, and different decay modes will change the n/p ratio in different ways.  Perhaps your question is why uranium undergoes alpha decay instead of one of the other decay modes.  That is a much more difficult question.

yes yes that's my question.never thought u were misunderstanding what i wanted ::)my poor English

There are several possible ways for radioactive nuclides to decay, most notably alpha decay, beta decay, electron capture and spontaneous fission:  Alpha decay involved the emission of a stripped 4He nucleus from the atom.  Beta decay involves the emission of an electron or a positron (along with neutrinos) from the nucleus.  During B+ decay a proton is converted into a neutron and the reverse is true for B- decay.  Electron capture is the process by which an orbital electron is absorbed in the nucleus, destroys a nuclear proton and creates a neutron.  Spontaneous fission involves the splitting of a nucleus into two or more smaller nuclei.

In order to simplify the ²³⁸U problem, let’s ignore the spontaneous fission decay mode and focus only on why ²³⁸U decays via alpha emission instead of beta decay or electron capture.

The Q value of a reaction determines whether or not a reaction is thermodynamically stable or not.  In the case of nuclear reactions, the Q value is simply the mass of the reactants minus the mass of the products (the reverse of most of chemistry).  A positive Q value implies that the reaction is exothermic and that over time the reaction will tend towards the products.  The reverse is true for a negative Q value.  In this case, the reaction is endothermic and over time will favor the reactants.

The values below are the Q values for each reaction:
Alpha decay: 4.27 MeV
B- decay: -0.147
B+ decay: -4.483
Electron Capture: -3.461

As you can see, the only decay mode that will release energy is alpha decay.  As the other modes all require energy, ²³⁸U will not decay by them.


On a fascinating side note, the Table of Radioactive Isotopes lists ²³⁸U as having a bb decay, which I assume means double beta decay.

Double beta decay is the simultaneous conversion of 2 neutrons into 2 protons under emission of 2 electrons and 2 anti neutrinos. The important part is that both conversions happen at the same time and not successively.
This type of decay has been postulated, and there are several large scale experiments planned to examine it. But as far as I know in the case of uranium there has only been a limit determined so far. One way of determining the rate for the double beta decay of uranium would be to measure the amount of plutonium 238 that is produced in a sample of uranium 238 over time. Unfortunately you have to make sure that your uranium does not contain any traces of plutonium from other sources. And this is very, very difficult.