<span>The number of the group identifies the column of the standard periodic table in which the element appears.</span>
Group 1 contains the alkali metals ( lithium<span> (</span>Li<span>), </span>sodium<span> (</span>Na<span>), </span>potassium<span> (</span>K<span>), </span>rubidium<span> (</span>Rb<span>), </span>caesium<span> (</span>Cs<span>), and </span>francium(Fr).)<span>
Group 2 contains the alkaline earth metals (</span> beryllium<span> (</span>Be),magnesium<span> (</span>Mg<span>), </span>calcium<span> (</span>Ca<span>), </span>strontium<span> (</span>Sr<span>), </span>barium<span> (</span>Ba<span>) and </span>radium<span> (</span>Ra<span>) )
Group 3: </span><span> Scandium (Sc) and yttrium (Y) </span>
Answer:
b) q large and m small
Explanation:
q is large and m is small
We'll express it as :
q > m
As we know the formula:
F = Eq
And we also know that :
F = Bqv
F = 
Bqv = 
or Eq = 
Assume that you want a velocity selector that will allow particles of velocity v⃗ to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.
When an object absorbs an amount of energy equal to Q, its temperature raises by

following the formula

where m is the mass of the object and

is the specific heat capacity of the material.
In our problem, we have

,

and

, so we can re-arrange the formula and substitute the numbers to find the specific heat capacity of the metal: