Can we find a unified theory of physics? .
Answers
Answer:
Solution :-
We Know,
\text m_ \alpha = 4\text{amu}m
α
=4amu
\text m_\text p = 1\text{amu}m
p
=1amu
❒ Calculating Kinetic Energies :
⏩ For α - particle :
\begin{gathered}\text{K.E}_\alpha = \frac{1}{2} \text m_ \alpha\text v_ \alpha {}^{2} \\ \end{gathered}
K.E
α
=
2
1
m
α
v
α
2
\begin{gathered}\text{K.E}_\alpha = \frac{1}{2} \times 4\times \text v_ \alpha {}^{2} \\ \end{gathered}
K.E
α
=
2
1
×4×v
α
2
\begin{gathered} \large\green{ \pmb{\bf{K.E}_\alpha = 2 v_ \alpha {}^{2}}} \\ \end{gathered}
K.E
α
=2v
α
2
K.E
α
=2v
α
2
⏩ For proton :
\begin{gathered}\text{K.E}_\text p = \frac{1}{2} \text m_ \text p\text v_ \text p {}^{2} \\ \end{gathered}
K.E
p
=
2
1
m
p
v
p
2
\begin{gathered}\text{K.E}_\text p = \frac{1}{2} \times 1\times \text v_ \text p {}^{2} \\ \end{gathered}
K.E
p
=
2
1
×1×v
p
2
\begin{gathered} \large\green{ \pmb{\bf{K.E}_\text p = \dfrac{1}{2} v_ \text p {}^{2}}} \\ \end{gathered}
K.E
p
=
2
1
v
p
2
K.E
p
=
2
1
v
p
2
As Given in Question kinetic energy is same of both α - particle and proton :
\begin{gathered} \dfrac{\text K.\text E_ \alpha }{\text K.\text E_\text p} = 1 \\ \end{gathered}
K.E
p
K.E
α
=1
\begin{gathered}:\longmapsto \frac{2 \times {\text v_ \alpha }^{2} }{ \frac{1}{2} \times\text v_\text p {}^{2} } = 1 \\ \end{gathered}
:⟼
2
1
×v
p
2
2×v
α
2
=1
\begin{gathered}:\longmapsto \text v_\text p {}^{2} = 4 \text v_ \alpha {}^{2} \\ \end{gathered}
:⟼v
p
2
=4v
α
2
\purple{ \Large :\longmapsto \underline {\boxed{{\pmb{\bf v_p = 2v_\alpha}} }}}:⟼
v
p
=2v
α
v
p
=2v
α
❒ We Have 3rd Equation of Motion as :
\pink{ \bigstar \: \: } \large \bf \orange{ \underbrace{ \underline{ {v}^{2} - {u}^{2} = 2as }}}★
v
2
−u
2
=2as
⏩ Applying 3rd Equation For α - particle
\begin{gathered}:\longmapsto\text v_ \alpha {}^{2} - \text u_ \alpha {}^{2} = 2\text a\text s_ \alpha \\ \end{gathered}
:⟼v
α
2
−u
α
2
=2as
α
As Initial Velocity is 0
\begin{gathered}:\longmapsto\text v_ \alpha {}^{2} - 0 = 2\text a\text s_ \alpha \\ \end{gathered}
:⟼v
α
2
−0=2as
α
\red{\large \pmb{ \bf 2as_ \alpha = v_ \alpha {}^{2} }} \: - - - (1)
2as
α
=v
α
2
2as
α
=v
α
2
−−−(1)
⏩ Applying 3rd Equation For proton :
\begin{gathered}:\longmapsto\text v_ \p {}^{2} - \text u_ \p {}^{2} = 2\text a\text s_ \p \\ \end{gathered}
:⟼v
\p
2
−u
\p
2
=2as
\p
As Initial Velocity is 0
\begin{gathered}:\longmapsto\text v_ \p {}^{2} - 0 = 2\text a\text s_ \p \\ \end{gathered}
:⟼v
\p
2
−0=2as
\p
\red{\large{ \bf 2as_ \p = v_ \p {}^{2} }} \: - - - (2)2as
\p
=v
\p
2
−−−(2)
⏩ Dividing (2) by (1) :
\begin{gathered}\pmb{\bf\dfrac{s_{\alpha}}{s_p} = \frac{ {v_ \alpha }^{2} }{v_p {}^{2} } } \\ \end{gathered}
s
p
s
α
=
v
p
2
v
α
2
s
p
s
α
=
v
p
2
v
α
2
Putting \pmb{\bf v_p = 2v_\alpha}
v
p
=2v
α
v
p
=2v
α
\begin{gathered}:\longmapsto \sf\dfrac{s_{\alpha}}{s_p} = \frac{ {v_ \alpha }^{2} }{(2v_ \alpha) {}^{2} } \\ \end{gathered}
:⟼
s
p
s
α
=
(2v
α
)
2
v
α
2
\begin{gathered}:\longmapsto \sf\dfrac{s_{\alpha}}{s_p} = \frac{{ \cancel{v_ \alpha }^{2}} }{4 \: \cancel{v_ \alpha {}^{2} }} \\ \end{gathered}
:⟼
s
p
s
α
=
4
v
α
2
v
α
2
\purple{ \Large :\longmapsto \underline {\boxed{{\pmb{\dfrac{s_{p}}{s_ \alpha } = \frac{4}{1} } }}}}:⟼
s
α
s
p
=
1
4
s
α
s
p
=
1
4
Explanation: