A new element has been produced called Volunteerium. This unstable isotope has 1

Question

A new element has been produced called Volunteerium. This unstable isotope has 104 protons and 162 neutrons. It decays by alpha emission to Nobelium. (SEE PICTURE. Need help with all parts. My answers are wrong)

A new element has been produced, called Volunteerium (Vo). This unstable isotope has 104 protons and 162 neutrons. It decays by alpha emission to Nobelium (No). Answer the three questions below, using three significant figures Part A: What is the atomic mass number (A) for the Nobelium that results from the decay of the Volunterium? Number A= Enter a whole number value.) rt B: Suppose that 112 micrograms of Vo are initially produced, after 6.00 seconds, there are 56.0 micrograms remaining. How long does it take (t) before there are only 14.0 micrograms remaining? Number Part C: What is the decay constant (A) for Vo? Number

Explanation / Answer

A.) Alpha particle is nothing but a Helium nucleus, which has 2 protons and 2 neutrons

So, 266104Vo -> 42He + 262102No

SO, as we see the mass number of Nobelium is A = 262

B.) Half of initial mass (112/2 = 56) is left after 6 seconds.

Which means the half life of the given element is 6 seconds

So, it takes 6 seconds to half the existing part and to be left with 56 /2 = 28 micrograms

So, after 12 seconds, we are left with 28 micrograms

And it takes another half life to half what is existing and be left with 28 /2 = 14 micrograms

So, after 12 + 6 = 18 seconds we are left only with 14 micrograms

Ans = 18 seconds

C.) the decay constant is given by = ln2 / half -life = 0.693 / T1/2  

= 0.693 / 6 = 0.1155 s-1

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