Question

Solve question no. 2 and 3, explain the ans step by step. Thanks!​

Answer 1

HOPE IT WILL HELP YOU.......

Answer 2

Answer:

1)Given :-

• In the Rutherford scattering experiment the number of alpha particles scattered at an angle of
• $theta = {60}^{0} \: is \: 12 \: per \: minute$

□To find :-

• The number of alpha particles per min when scattered at an angle of 90 degree is what?

□Explanation :-

• The scattered alpha particles unit area at an angle of theta is proportional to
• $\frac{1}{ {sin}^{4} } \times \frac{theta}{2}$
• $\frac{N2}{N1} = ( \frac{ {( \frac{sin \: theta}{2} }^{4} ))}{ \frac{sintheta2}{2} } )$
• $\frac{n2}{36} = ( \frac{ \frac{ {sin60}^{4} }{2} }{ \frac{sin \: 90}{2} } )$
• $\frac{n2}{36} = \frac{1}{4}$
• $N2 = \frac{36}{4}$
• $N2 = 9$

♧Therefore , the answer for 1st question is N2=9.

□2)Explanation: -

• Here the energy required to break one Cl -Cl bond
• $= \frac{bond \: energy \: \: per \: mole}{avagadro \: number}$
• $= \frac{119.255 \times {10}^{3} }{6.023 \times {10}^{23} } j$

□Let,

• Wavelength of photon which is required to break one Cl-Cl bond be lambda.
• $lambda = \frac{hc}{E}$
• Now applying the values we get that,

• $= \frac{119.255 \times {10}^{ - 34 } \times {10}^{31 } \times {10}^{ - 3} }{243}$
• $= 4.91 \times {10}^{ - 7}m$

♧Hope it helps u @Abhi4you.

♧Thank you .