Question

Two particle A and B are in motion. If the wavelength associated with 'A' is 33.33 nm, the wavelength associated with 'B' whose momentum is $\frac{1}{3}rd$ of 'A' is

• A. $1.0\times {10}^{-8}m$
• B. $2.5\times {10}^{-8}m$
• C. $1.25\times {10}^{-7}m$
• D. $1.0\times {10}^{-7}m$

Answer 1

The correct option is D $1.0\times {10}^{-7}m$

$(mV{)}_A=\frac{h}{{\lambda }_A}$
$(mV{)}_B=\frac{h}{{\lambda }_B}$
$(mV{)}_B=\frac{1}{3}(mV{)}_A$
$\frac{h}{{\lambda }_B}=\frac{h}{3{\lambda }_A}$
$=2\times 5\times {10}^{-8}m$
$=10\times {10}^{-7}m$