Question

In a nuclear physics experiment, a proton is fired toward a Z = 13 nucleus with the diameter and neutron energy levels shown in Figure 40.17. The nucleus, which was initially in its ground state, subsequently emits a gamma ray with wavelength 1.73×10−4 nm. What was the minimum initial speed of the proton? Hint: Don't neglect the proton-nucleus collision.

Accepted Answer

Understand the problem: The proton collides with the nucleus, exciting it to a higher energy state. The nucleus then emits a gamma ray as it transitions back to its ground state. The goal is to find the minimum initial speed of the proton required to excite the nucleus. This involves energy conservation and quantum mechanics principles. Step 1: Calculate the energy of the emitted gamma ray. Use the formula for the energy of a photon: $$ E = \frac{hc}{\lambda} $$, where $$ h  is $$Planck's constant ($$ 6.626 	imes 10^{-34} \ 	ext{J·s} $$), $$ c  is $$the speed of light ($$ 3.00 	imes 10^8 \ 	ext{m/s} $$), and $$ \lambda  is $$the wavelength of the gamma ray ($$ 1.73 	imes 10^{-4} \ 	ext{nm} $$, converted to meters). Step 2: Recognize that the energy of the gamma ray corresponds to the energy difference between the excited state and the ground state of the nucleus. This means the proton must transfer at least this much energy to the nucleus during the collision to excite it. Step 3: Use the principle of conservation of energy. The kinetic energy of the proton before the collision is given by $$ KE = \frac{1}{2}mv^2 $$, where $$ m  is $$the mass of the proton ($$ 1.67 	imes 10^{-27} \ 	ext{kg} ) $$and $$ v  is $$its speed. Set this equal to the energy of the gamma ray to find the minimum speed of the proton: $$ \frac{1}{2}mv^2 = E . $$Step 4: Solve for $$ v  by $$rearranging the equation: $$ v = \sqrt{\frac{2E}{m}} . $$Substitute the values for $$ E  ($$calculated in Step 1) and $$ m  to $$find the minimum initial speed of the proton.

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Nuclear Reactions

Energy Conservation

Wavelength and Energy Relationship