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P

phuong30101993

N

newstarinsky

ta có

[TEX]hc.(\frac{1}{\lambda_1}-\frac{1}{\lambda_0})=\frac{1}{2}.mv^2_1(1)[/TEX]

có [TEX]\frac{\lambda_1}{\lambda_2}=\frac{6}{3}\Rightarrow \lambda_2=\frac{1}{2}.\lambda_1[/TEX]

[TEX]hc.(\frac{2}{\lambda_1}-\frac{1}{\lambda_0})=\frac{1}{2}.9mv^2_1(2)[/TEX]

lấy (2) chia cho (1) ta được

[TEX]\lambda_o=\frac{8}{7}\lambda_1[/TEX]

nên (1)trở thành

[TEX]\frac{1}{8\lambda_1}hc=\frac{1}{2}mv^2_1(3)[/TEX]

lại có

[TEX]hc.(\frac{1}{\lambda_3}-\frac{1}{\lambda_o})=\frac{1}{2}mk^2v^2_1[/TEX]

[TEX]\Leftrightarrow hc.(\frac{3}{2\lambda_1}-\frac{7}{8\lambda_1})=\frac{1}{2}mk^2v^2_1[/TEX]

[TEX]\Leftrightarrow \frac{5}{8\lambda_1}hc=\frac{1}{2}mk^2v^2_1(4)[/TEX]

Lấy (4) chia cho (3) ta được

[TEX]k^2=5\Rightarrow k=\sqrt{5}[/TEX]
 
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