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Theo bài ra ta có
ε K N = E 4 - E 1 = -13,6/16 - (-13,6/1) = 12,75eV
λ = hc/ ε = 0,9742. 10 - 7 m = 0,0974 μ m ⇒ thuộc vùng tử ngoại.
\(_1^1p + _3^7 Li \rightarrow 2_2^4He\)
\(\Delta m = (m_p+m_{Li}- 2m_{He}) = 0,0187u>0 \)
=> \(m_t > m_s \), phản ứng tỏa năng lượng.
\(E = \Delta m c^2= 0,0187.931 =17,4097 MeV.\)
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\(_1^1p + _3^7 Li \rightarrow _2^4He+_2^4He\)
\(W_{tỏa} = (m_t-m_s)c^2 =( m_{Li}+m_p - 2m_{He}).931=17,4097MeV.\)
Số hạt nhân \(_2^4He\) trong 1,5 g heli là \(N= nN_A= \frac{m}{A}.N_A = \frac{1,5}{4}.6,02.10^{23}= 2,2575.10^{23} \)(hạt)
Mỗi phản ứng tạo ra 2 hạt nhân \(_2^4He\) thì tỏa ra năng lượng là 17,4097 MeV
=> Để tạo ra 2,2572.1023 hạt nhân \(_2^4He\) thì tỏa ra năng lượng là
\(W = \frac{17,4097.2,2575.10^{23}}{2} = 1,965.10^{24}MeV.\)
