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\(_1^1p + _3^7 Li \rightarrow 2_2^4He\)
Phản ứng tỏa năng lượng nên \(W_{tỏa} = (m_t-m_s)c^2 = 2K_{He}-(K_p+K_{Li})\)
=> \( 2K_{He} = (m_p+m_{Li}-2m_{He})c^2+ K_p\) (do Li đứng yên nên KLi = 0)
=> \(K_{He} = 9,6 MeV = 9,6.10^6.1,6.10^{-19}J.\)
=> \(v = \sqrt{\frac{2K_{He}}{m_{He}}} = \sqrt{\frac{2.9,6.10^6.1,6.10^{-19}}{4,0015.1,66.10^{-27}}} = 21505282,4 m/s.\)
Năng lượng toàn phần trong quá trình phản ứng hạt nhân xảy ra được bảo toàn
Lực lorenxo tác dụng lên hạt α khi nó chuyển động trong từ trường
Đáp án B
\(_1^1p + _3^7 Li \rightarrow 2_2^4He\)
Phản ứng là tỏa năng lượng nên
\(W_{tỏa} = (m_t-m_s)c^2 = K_s-K_t\)
=> \(m_p +m_{Li} - 2m_{He} =2K_{He} - K_p\) (do Li đứng yên nên KLi = 0)
=> \(2K_{He} = K_p+(m_p+m_{Li}-2m_{He})c^2 = 1,8 + 0,0187.931 = 19,2097MeV\)
=> \(K_{He} = 9,6 0485 MeV.\)
Q=(mt-ms).931 MeV= -1.21 MeV
mà Q=Ks-Kt >> -1.21=Kp+Kx-4 >> Kp+Kx=2.79
>> 1/2MxVx+1/2MpVp=2.79
mà Vp=Vx >> 1/2Vp(Mp+Mx)=2.79 >> Vp=0.5.10^7m/s >> Kp=0.1306MeV
\(_1^1p + _3^7 Li \rightarrow 2_2^4He\)
\(\Delta m = (m_p+m_{Li}- 2m_{He}) = 0,0187u>0 \)
=> \(m_t > m_s \), phản ứng tỏa năng lượng.
\(E = \Delta m c^2= 0,0187.931 =17,4097 MeV.\)
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Đáp án D