lander=1.2*10^12
muon mass=207*9.11*10^-31
muon speed=3*10^6
neutral pion mass=134.972

calculate the speed necessary for a (neutral pion) to have the same de Broglie wavelength as that of the muon (lander)

the mark scheme uses the conservation of momentum, MV(of pion)=MV(of muon)

yet it ends up with (3*10^6)/1.28 (1.28 is the ratio (rest mass pion)/(rest mass muon))

I don't understand how it gets this please help quick

Please any help would be greatly appreciated

1. It's lambda not lander

2. i believe it's using \lambda = \frac{h}{mv} (De Broglie Hypothesis) and NOT the conversation of mass (there's no reason to use that, the question doesn't ask for it).

since \lambda is the same for both particles, \frac{h}{m_{\mu}v_{\mu}} = \frac{h}{m_{\pi}v_{\pi}} and therefore m_{\mu}v_{\mu} = m_{\pi}v_{\pi}

it's not conversation of momentum.

rearranging \frac{m_{\mu}}{m_{\pi}}v_{\mu} = v_{\pi}

which, when you put the number in, I believe is what the markscheme gives as the answer.

Please ask if you don't understand.

the mark scheme gets 2.3*10^6m/s

I get1.7*10^4
what do u get

(but it does use MuVu=MpiVpi

\frac{m_{\mu}}{m_{\pi}}v_{\mu} = \frac{105}{134}3E6 \approx 2.3E6ms^{-1}

I seem to agree with the mark scheme.

Damm my matchs was wrong thanks for helping (I would rate u but still cant)