1. The problem statement, all variables and given/known data Compare 25-MeV alpha particles wavelengths with the size of a nucleon in a nucleus. 2. Relevant equations λ = h/p E2 = p2c2 + m2c4 KE = ½mv2 p = mv 3. The attempt at a solution To compare the wavelength of an α particle with a nucleon, I would need to divide the wavelength of this 25-MeV α particle by the size of a nucleon, 1×10-15 (I looked this value up). So I'm looking for λα / dnucleon. First I solved for p using the second relevant equation, which gave me: p = √((E2 - (mc2)2)/c2). E = 25 MeV, or the energy of the alpha particle, and mc2 = 4.00150618 u⋅c2 × 931.5 MeV/c2 = 3727.403 MeV. So p = √(((25 MeV)2 - (3727.403 MeV)2)/c2). This, however, gives me a negative number under the square root. I then tried to use the third and fourth relevant equations to get p. The mass of the α particle in kilograms is 6.64465675×10-27. I converted the α particle's energy from MeV to joules: 25 MeV = 4.0054 × 10-12 J = 4.0054 × 10-12 kg⋅m2/s2 Solving for v gave me: v = √(2(4.0054×10-12 kg⋅m2/s2) / 6.64465675×10-27 kg) v = 34721755.06 m/s p = mv p = (6.64465675×10-27 kg)(34721755.06 m/s) = 2.30714114×10-19 Plugging in p for the wavelength formula gives me: λ = h/p = (6.626×10-34 kg⋅m2/s) / (2.30714114×10-19 kg⋅m/s) λ = 2.87195223×10-15 m Finally, I divide λ by the size of a nucleon, which would be 1×10-15. λα / dnucleon = 2.87195223×10-15 m / 1×10-15 m λα / dnucleon = 2.87195 However, this answer is incorrect. The actual answer at two significant figures is 1.2, and I can't for the life of me figure out what I did wrong. I have a second problem exactly like this one, but with a proton particle instead, so I'm hoping that figuring out what I did wrong on this problem will help me solve the second problem. If anyone can help me, it would be much appreciated!