As per the given question,
Applied potential (V)=450V
Charge on electron ( e ) = 1.6 × 1 0 − 6 C (e)=1.6\times 10^{-6}C ( e ) = 1.6 × 1 0 − 6 C
B = 1.5 × 1 0 − 3 T B=1.5 \times 10^{-3}T B = 1.5 × 1 0 − 3 T
We know that,
q v × B = m v 2 R qv\times B=\frac{mv^2}{R} q v × B = R m v 2
⇒ q v × B = m v 2 R \Rightarrow qv\times B=\frac{mv^2}{R} ⇒ q v × B = R m v 2
⇒ R = 1 q B 2 m 2 v 2 2 \Rightarrow R=\frac{1}{qB}\sqrt{\frac{2m^2 v^2}{2}} ⇒ R = qB 1 2 2 m 2 v 2
⇒ q v × B = 2 e V R \Rightarrow qv\times B=\frac{2eV}{R} ⇒ q v × B = R 2 e V
⇒ R = 2 m e V e × B \Rightarrow R=\frac{\sqrt{2 m_e V}}{\sqrt{e}\times B} ⇒ R = e × B 2 m e V
⇒ R = 2 × 9.1 × 1 0 − 31 × 450 1.6 × 1 0 − 19 × 1.5 × 1 0 − 3 = 8190 × 1 0 − 31 1.89 × 1 0 − 3 \Rightarrow R = \frac{\sqrt{2\times 9.1 \times 10^{-31}\times 450}}{\sqrt{1.6\times 10^{-19}}\times 1.5\times 10^{-3}}=\frac{\sqrt{8190\times 10^{-31}}}{1.89\times 10^{-3}} ⇒ R = 1.6 × 1 0 − 19 × 1.5 × 1 0 − 3 2 × 9.1 × 1 0 − 31 × 450 = 1.89 × 1 0 − 3 8190 × 1 0 − 31
⇒ R = 28.61 × 1 0 − 15 1.89 × 1 0 − 3 m \Rightarrow R=\frac{28.61\times 10^{-15}}{1.89\times 10^{-3}}m ⇒ R = 1.89 × 1 0 − 3 28.61 × 1 0 − 15 m
⇒ R = 15.13 × 1 0 − 12 m \Rightarrow R = 15.13\times 10^{-12} m ⇒ R = 15.13 × 1 0 − 12 m
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