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  • Two particles A1 and A2 of masses m1, m2 in to bracket m1 grater m2 have the same de Broglie wavelength. Then (a) their momenta are the same (b) their energies are the same (c) energy of A1 is less than the energy of A2 (d) energy of A1 is more than the

Two particles A1 and A2 of masses m1, m2 (m1 > m2) have the same de Broglie wavelength. Then

(a) their momenta are the same

(b) their energies are the same

(c) energy of A1 is less than the energy of A2

(d) energy of A1 is more than the energy of A2

Answers (1)

The answer is the option (a. c)

\lambda=\frac{h}{p}\\ p=\frac{h}{\lambda}\;or\; p\alpha \frac{1}{\lambda}\\ \frac{p_{1}}{p_{2}}=\frac{\lambda_{2}}{\lambda_{1}}\\ \lambda_{1}=\lambda_{2}=\lambda (given) p_{1}=p_{2} \; vertifies(a)\\ E_{a}=\frac{1}{2}mv^{2}=\frac{\frac{1}{2}m^{2}v^{2}}{m}=\frac{p^{2}}{2m}\\ E\alpha \frac{1}{m}[as\;p_{1}=p_{2}]\\ \frac{E_{1}}{E_{2}}=\frac{m_{2}}{m_{1}}\\ \frac{E_{1}}{E_{2}}<1[m_{1}>m_{2}] E_{2}>E_{1}

 Thus c is correct.
 

Posted by

infoexpert24

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