Question

chem 1123
chapter 3: electromagnetic spectrum and
periodic properties
the de broglie wavelength is a characteristic of particles, not electromagnetic radiation.
particles have wave - like behavior and waves have particle - like behavior. there
is no clear distinction at the microscopic level.
problem:
if an electron travels at a velocity of 1.000×10⁷ ms⁻¹ and has a mass of 9.109×10⁻²⁸ g, what is its
wavelength? $\lambda = \frac{h}{mv} = \frac{h}{p}$
what is wave - particle duality?
what is the heisenberg uncertainty principle?
schrödinger thought of the electron in terms of a three - dimensional stationary wave, or
________, represented by the greek letter psi ____.
max born claimed that wavefunctions are ____________________ amplitudes.
what do we use wavefunctions to determine?
quantum mechanics is a ____________________ theory in science that provides description of
the physical properties of nature at the scale of atoms and subatomic particles.
the energy of an electron in an atom is ______________.
do electrons smoothly transition between two energy levels?

Explanation:

Step1: Recall de Broglie wavelength formula

The de Broglie wavelength formula is \(\lambda=\frac{h}{p}\), where \(p = mv\) (momentum), \(h = 6.626\times10^{-34}\space J\cdot s\) (Planck's constant), \(m\) is mass, and \(v\) is velocity. First, convert the mass of the electron from grams to kilograms: \(m = 9.109\times10^{-28}\space g=9.109\times10^{-31}\space kg\) (since \(1\space kg = 1000\space g\)).

Step2: Calculate momentum \(p\)

\(p=mv\), substitute \(m = 9.109\times10^{-31}\space kg\) and \(v = 1.000\times10^{7}\space m/s\). So \(p=(9.109\times10^{-31}\space kg)\times(1.000\times10^{7}\space m/s)=9.109\times10^{-24}\space kg\cdot m/s\).

Step3: Calculate wavelength \(\lambda\)

Using \(\lambda=\frac{h}{p}\), substitute \(h = 6.626\times10^{-34}\space J\cdot s\) and \(p = 9.109\times10^{-24}\space kg\cdot m/s\). \(\lambda=\frac{6.626\times10^{-34}}{9.109\times10^{-24}}\approx7.274\times10^{-11}\space m\) (or \(72.74\space pm\), since \(1\space pm = 10^{-12}\space m\)).

Answer:

The wavelength of the electron is approximately \(\boldsymbol{7.27\times10^{-11}\space m}\) (or \(72.7\space pm\)).