Q.The threshold frequency for a certain metal is $3.3 \times 10^{14}\ \text{Hz}$. If light of frequency $8.2 \times 10^{14}\ \text{Hz}$ is incident on the metal, predict the cut-off voltage for the photoelectric emission.

Concept and Intuition

Einstein's photoelectric equation states that the maximum kinetic energy of an emitted photoelectron equals the photon's energy minus the metal's work function:

The stopping potential $V_0$ is the retarding voltage that just brings the fastest photoelectrons to rest, so $eV_0 = KE_{max}$, giving $V_0$ numerically in volts when energies are expressed in eV.

Step-by-Step Solution

1. Photon energy using the handy constant $hc \approx 12400\ eV\cdot\text{Å}$: $E = \dfrac{12400}{2000} = 6.2$ eV.
2. Work function given: $\phi = 5.01$ eV.
3. Maximum kinetic energy: $KE_{max} = E - \phi = 6.2 - 5.01 = 1.19$ eV. …

Concept and Intuition

In the photoelectric effect, a single photon transfers all its energy to one electron. Part of that energy overcomes the metal's work function $\phi$ (the minimum energy needed to free the electron), and whatever is left over becomes the electron's kinetic energy — this is Einstein's photoelectric equation, $KE_{max}=E_{photon}-\phi$.

Step-by-Step Solution

1. Convert wavelength: $1000\,\text{Å}=1000\times10^{-10}\,\text{m}=100\,\text{nm}$. …

Concept and Intuition

The maximum kinetic energy of photoelectrons is $KE_{max} = h\nu - \phi$, where $h\nu$ is the photon energy and $\phi$ is the work function. Momentum and kinetic energy for a non-relativistic electron are related by $p = \sqrt{2mKE}$, so momentum scales as the square root of kinetic energy, not linearly with photon energy.

Step-by-Step Solution

1. First case: $KE_1 = 4.2\,\text{eV} - 2.2\,\text{eV} = 2.0\,\text{eV}$, with momentum $p_1 = \sqrt{2mKE_1} = P$.
2. Second case: $KE_2 = 6.2\,\text{eV} - 2.2\,\text{eV} = 4.0\,\text{eV} = 2 \times KE_1$.
3. Momentum ratio: $\dfrac{p_2}{p_1} = \sqrt{\dfrac{KE_2}{KE_1}} = \sqrt{2}$.
4. So $p_2 = \sqrt{2}\,P$. …

Concept and Intuition

Einstein's photoelectric equation says the maximum kinetic energy of an emitted photoelectron equals the incident photon's energy minus the material's work function: $KE_{max}=\dfrac{hc}{\lambda}-\phi$. The problem first gives us a calibration point (2500 Å ↔ 4.96 eV) so we don't need to remember $hc$ precisely — we can find the effective $hc$ product from it.

Step-by-Step Solution

1. From the given data: $E=\dfrac{hc}{\lambda}\Rightarrow hc = 4.96\,eV\times2500\,Å=12400\,eV\cdot Å$ — this matches the well-known constant $hc\approx12400\,eV\cdot Å$.
2. Photon energy at 3100 Å: $E'=\dfrac{12400}{3100}=4.0\,eV$.
3. Maximum kinetic energy of photoelectron: $KE_{max}=E'-\phi=4.0-2.2=1.8\,eV$.
4. Convert to joules: $KE_{max}=1.8\times1.6\times10^{-19}=2.88\times10^{-19}\,J$. …

Concept and Intuition

The photoelectric effect: each photon transfers its entire energy $hc/\lambda$ to one electron; the electron escapes only after "paying" the work function, with any leftover energy becoming kinetic energy.

Step-by-Step Solution

1. Photon energy: $E=\dfrac{hc}{\lambda}=\dfrac{1240\ \text{eV nm}}{200\ \text{nm}}=6.2$ eV.
2. Maximum kinetic energy: $KE_{max}=E-\phi=6.2-4.2=2.0$ eV $=2.0\times1.6\times10^{-19}=3.2\times10^{-19}$ J. …

Concept and Intuition

Einstein's photoelectric equation gives the maximum kinetic energy of an emitted electron as $KE = hc/\lambda - hc/\lambda_0 = hc\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)$, where $\lambda_0$ is the threshold wavelength. Since $KE = \tfrac12 mv^2$, the velocity ratio is the square root of the kinetic energy ratio.

Step-by-Step Solution

1. For $\lambda_1 = \lambda_0/3$: $KE_1 = hc\left(\dfrac{3}{\lambda_0}-\dfrac{1}{\lambda_0}\right) = \dfrac{2hc}{\lambda_0}$.
2. For $\lambda_2 = \lambda_0/9$: $KE_2 = hc\left(\dfrac{9}{\lambda_0}-\dfrac{1}{\lambda_0}\right) = \dfrac{8hc}{\lambda_0}$.
3. $KE_1:KE_2 = 2:8 = 1:4$. …

Concept and Intuition

Einstein's photoelectric equation states $\frac12 m v_{max}^2 = \frac{hc}{\lambda} - W$, where $W$ is the work function. Given two different illuminating wavelengths, we get two equations in two unknowns ($W$ and the electron's mass/velocity scale); the given velocity ratio lets us eliminate the common kinetic-energy scale factor and solve directly for $W$.

Step-by-Step Solution

1. Let $K_1 = \frac12 m v_1^2$ (for $\lambda_1=300$ nm) and $K_2 = \frac12 m v_2^2$ (for $\lambda_2=500$ nm). Given $v_1/v_2 = 3 \Rightarrow K_1/K_2 = 9$, so $K_1 = 9K_2$.
2. Einstein's equations: $9K_2 = \dfrac{hc}{\lambda_1} - W$ and $K_2 = \dfrac{hc}{\lambda_2} - W$.
3. Subtract: $8K_2 = \dfrac{hc}{\lambda_1} - \dfrac{hc}{\lambda_2}$. …

Concept and Intuition

Einstein's photoelectric equation, $KE_{max}=hf_{incident}-\phi$, lets us relate the maximum kinetic energy (hence maximum velocity) of photoelectrons to the incident frequency and the material's work function. Given data at one frequency lets us solve for the unknown constant $hf$, which can then be used at the new frequency.

Step-by-Step Solution

1. At frequency $3f$: $\tfrac12mv^2 = h(3f) - 2hf = hf$.
2. So $hf = \tfrac12 mv^2$ — this is a useful relation we'll reuse.
3. At frequency $4.25f$: $KE' = h(4.25f) - 2hf = 2.25hf$. …

Concept and Intuition

Only photons with energy above the metal's work function can eject electrons; the excess energy $hf-\phi$ becomes the electron's kinetic energy. This is the foundational evidence for the particle (photon) nature of light.

Step-by-Step Solution

1. Photon energy: $hf = 6.6\times10^{-34}\times5\times10^{14} = 3.3\times10^{-19}$ J.
2. Work function: $\phi = 2.0\,\text{eV} = 2.0\times1.6\times10^{-19} = 3.2\times10^{-19}$ J.
3. Maximum KE: $KE_{max} = 3.3\times10^{-19} - 3.2\times10^{-19} = 1.0\times10^{-20}$ J.
4. From $KE_{max}=\tfrac12 m v_{max}^2$: $v_{max} = \sqrt{\dfrac{2\times1.0\times10^{-20}}{9\times10^{-31}}} = \sqrt{2.222\times10^{10}}$.
5. $v_{max} \approx 1.4907\times10^5$ m/s. …

Concept and Intuition

When photoelectrons enter a magnetic field, they move in circles of radius $r=mv/(eB)$ (magnetic force provides centripetal force). Since $v$ varies (electrons are ejected with a range of kinetic energies up to a maximum, set by Einstein's equation $KE_{max}=hf-\phi$), the largest radius corresponds to the fastest electrons — i.e. the maximum kinetic energy. This lets us work backward from the given radius to $KE_{max}$, and then to the work function.

Step-by-Step Solution

1. From $r = \dfrac{mv}{eB}$: $v = \dfrac{eBr}{m}$.
2. $KE_{max} = \dfrac12 mv^2 = \dfrac12 m\left(\dfrac{eBr}{m}\right)^2 = \dfrac{e^2B^2r^2}{2m}$.
3. Plug in: $e=1.6\times10^{-19}$ C, $B=2\times10^{-4}$ T, $r=0.03$ m, $m=9\times10^{-31}$ kg.
4. $e^2B^2r^2 = (2.56\times10^{-38})(4\times10^{-8})(9\times10^{-4}) = 9.216\times10^{-49}$. …