Question

calculate the accelerating potential that must be imparted to a proton beam to give it an effective wavelength of 0.005 nm

Answer 1

HELLO FRIEND HERE IS YOUR ANSWER,,,,,,,,


We can divide the following into two major steps to obtain the final Accelerating potential that is "V".


STEP 1: Calculating the velocity of those protons in that proton beam;

Here, The Mass of the Proton $\simeq$ to that of the Mass of Hydrogen = $\frac{1.008}{6.022 \times 10^{23}}$ Gram

$\therefore$ Mass of Hydrogen = $1.67 \times 10^{-24} \: Grams$
= $1.67 \times 10^{-27} \: Kilograms$

$\therefore$ Wavelength ($\lambda$) will be =

$0.005 \: nanometres \\ \\ \\ \lambda = 0.005 \times 10^{-9} \: metre \\ \\ \\ \lambda = 5 \times 10^{-12} \: metre$

By applying the Most used De-Broglie's wave equation,

$\lambda = \frac{h}{mv}$

Or, we can re-frame this again in terms of "v" to get,,

$v = \frac{h}{m \lambda}$

By substituting those following values into the given equation;

$h = 6.626 \times 10^{-34} kg \: m^2 \: s^{-1}$, $m = 1.67 \times 10^{-27} \: kg$ and $\lambda = 5 \times 10^{-12} \: m$

$\therefore \: \: v = \frac{6.626 \times 10^{-34} kg \: m^2 \: s^{-1}}{(1.67 \times 10^{-27} \: kg) \times (5 \times 10^{-12} \: m)}$

$\therefore \: \: v = 7.94 \times 10^4 \: m \: s^{-1}$



STEP 2: Now Calculating the required Accelerating potential by those given values;

If a Accelerating potential is represented as "V volts", then we will have that energy which is acquired by that proton as "eV", where "e" is going to denote the charge present on that specific proton (which will be equal to that charge on its electron). Then this becomes the kinetic energy of that proton.

Hence we are going to obtain,,,

$e \times V = \frac{1}{2} \: m \: v^2 \\ \\ OR \\ \\ \\ V = \frac{mv^2}{2 \times e}$

By substituting the give values,, that is;

$m = 1.67 \times 10^{-27} \: kg$, $v = 7.94 \times 10^4 \: m \: s^{-1}$ and $e = 1.602 \times 10^-19 \: Coulomb$

$\therefore \: \: V = \frac{(1.67 \times 10^{-27}) \times (7.94 \times 10^4)^2}{2 \times (1.602 \times 10^{-19})}$

$\therefore \: \: V = \frac{10^{-27} \times 1.67(10^4 \times 7.94)^2}{10^{-19} \times 3.204}$

$\therefore \: \: V = \frac{10^8 \times 1.67 \times 7.94^2}{10^8 \times 3.204}$

$\therefore \: \: V = \frac{1.67 \times 7.94^2}{3.204}$

$\therefore \: \: V = \frac{105.282812}{3.204}$

$\therefore \: \: V = 32.85980 \: (kg \: m^2 s^{-2}) \: (C^{-1}) \\ \\ \\ \therefore \: \: V = 32.85980 \: J \: C^{-1} \\ \\ \\ \therefore \: \: V = 32.85980 \: (CV) \: C^{-1} \\ \\ \\ \therefore \: \: V = 32.85980 \: V$

$\textbf{\huge{FINAL ANSWER : 32.86 \: V}}$


HOPE IT HELPS YOU AND SOLVES YOUR DOUBTS FOR THIS KIND OF QUERY TO FIND THE FINAL ACCELERATING POTENTIAL!!!!!!!

Answer 2

Answer:

Hope it helps

Explanation:

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