Chỉ cách làm câu 72 zới

Đáp án:

 B

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
\dfrac{{hc}}{{{\lambda _1}}} - \dfrac{{hc}}{{{\lambda _0}}} = \dfrac{1}{2}mv_1^2\\
\dfrac{{hc}}{{{\lambda _2}}} - \dfrac{{hc}}{{{\lambda _0}}} = \dfrac{1}{2}mv_2^2\\
 \Rightarrow \dfrac{{\dfrac{1}{{{\lambda _1}}} - \dfrac{1}{{{\lambda _0}}}}}{{\dfrac{1}{{{\lambda _2}}} - \dfrac{1}{{{\lambda _0}}}}} = {\left( {\dfrac{{{v_1}}}{{{v_2}}}} \right)^2}\\
 \Rightarrow \dfrac{{\dfrac{1}{{0,6}} - \dfrac{1}{{{\lambda _0}}}}}{{\dfrac{1}{{0,3}} - \dfrac{1}{{{\lambda _0}}}}} = {\left( {\dfrac{2}{4}} \right)^2}\\
 \Rightarrow {\lambda _0} = 0,9\mu m
\end{array}\)

Mặt khác:

\(\begin{array}{l}
\dfrac{{hc}}{{{\lambda _3}}} - \dfrac{{hc}}{{{\lambda _0}}} = \dfrac{1}{2}mv_3^2\\
 \Rightarrow \dfrac{{\dfrac{1}{{{\lambda _1}}} - \dfrac{1}{{{\lambda _0}}}}}{{\dfrac{1}{{{\lambda _3}}} - \dfrac{1}{{{\lambda _0}}}}} = {\left( {\dfrac{{{v_1}}}{{{v_3}}}} \right)^2}\\
 \Rightarrow \dfrac{{\dfrac{1}{{0,6}} - \dfrac{1}{{0,9}}}}{{\dfrac{1}{{0,2}} - \dfrac{1}{{0,9}}}} = {\left( {\dfrac{{{{2.10}^5}}}{{{v_3}}}} \right)^2}\\
 \Rightarrow {v_3} = 2\sqrt 7 {.10^5}\left( {m/s} \right)
\end{array}\)