Problem Solving Modern Physics

Question 1#

A photon of wavelength 300 nm strikes a metal surface with a work function of 2.0 eV. What is the kinetic energy of the most energetic electron emitted from the surface? (Given: $h = 6.63 \times 10^{-34} , \text{Js}$, $c = 3 \times 10^8 , \text{m/s}$, $1 , \text{eV} = 1.6 \times 10^{-19} , \text{J}$)

(1) 0.13 eV (2) 2.13 eV (3) 4.13 eV (4) 6.13 eV

Solution:

This problem involves the photoelectric effect. The energy of the incident photon ($E$) is given by:

$E = h\nu = \frac{hc}{\lambda}$

Where: $h$ = Planck’s constant = $6.63 \times 10^{-34} , \text{Js}$ $c$ = speed of light = $3 \times 10^8 , \text{m/s}$ $\lambda$ = wavelength of the photon = 300 nm = $300 \times 10^{-9} , \text{m}$

Substituting the values:

$E = \frac{(6.63 \times 10^{-34} , \text{Js}) \times (3 \times 10^8 , \text{m/s})}{300 \times 10^{-9} , \text{m}}$ $E = \frac{19.89 \times 10^{-26}}{3 \times 10^{-7}} , \text{J}$ $E = 6.63 \times 10^{-19} , \text{J}$

Now, we need to convert this energy to electron volts (eV):

$E (\text{in eV}) = \frac{6.63 \times 10^{-19} , \text{J}}{1.6 \times 10^{-19} , \text{J/eV}} \approx 4.14 , \text{eV}$

According to Einstein’s photoelectric equation, the kinetic energy of the most energetic emitted electron ($K_{max}$) is given by:

$K_{max} = E - \phi$

Where $\phi$ is the work function of the metal = 2.0 eV.

$K_{max} = 4.14 , \text{eV} - 2.0 , \text{eV} = 2.14 , \text{eV}$

The closest option is 2.13 eV.

Answer: (2)

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Question 2#

A radioactive nucleus has a half-life of 10 days. What fraction of the original number of nuclei will remain after 30 days?

(1) 1/2 (2) 1/4 (3) 1/8 (4) 1/16

Solution:

The number of nuclei remaining after time $t$ is given by:

$N(t) = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}$

Where: $N(t)$ = number of nuclei remaining after time $t$ $N_0$ = initial number of nuclei $t$ = total time = 30 days $T_{1/2}$ = half-life = 10 days

Substituting the values:

$N(30) = N_0 \left(\frac{1}{2}\right)^{30/10}$ $N(30) = N_0 \left(\frac{1}{2}\right)^{3}$ $N(30) = N_0 \times \frac{1}{2 \times 2 \times 2}$ $N(30) = N_0 \times \frac{1}{8}$

The fraction of the original number of nuclei remaining after 30 days is $\frac{N(30)}{N_0} = \frac{1}{8}$.

Answer: (3)