Wavelengths belonging to Balmer series lying in the range of 450 nm to 750 nm were used to eject pho

Q: Why do we only consider transitions to $n=2$?

The problem specifically mentions the 'Balmer series', which by definition consists of spectral lines corresponding to electron transitions from higher energy levels ($n > 2$) down to the second energy level ($n=2$).

Q: Why can't we just use the $450\text{ nm}$ wavelength directly to find the maximum energy?

The $450\text{ nm}$ wavelength is just the lower bound of the *filter* or *range* provided. The actual photons present are only those emitted by the hydrogen atom (Balmer series). We must check if a Balmer line actually exists at or near $450\text{ nm}$.

Q: What happens if a photon has energy less than the work function?

If the photon's energy is less than the work function ($2.0\text{ eV}$ in this case), it doesn't have enough energy to overcome the binding energy of the electron. No photoelectric emission will occur for that specific photon.

Q: Why did we reject the $n=5 \rightarrow 2$ transition?

The energy of the photon emitted in the $n=5 \rightarrow 2$ transition is $2.856\text{ eV}$. This corresponds to a wavelength shorter than $450\text{ nm}$, which falls outside the specified range of $450\text{ nm}$ to $750\text{ nm}$.

Q: Is the kinetic energy calculated the exact energy of every emitted electron?

No, it is the *maximum* possible kinetic energy ($K_{\max}$). Electrons can be emitted with kinetic energies ranging from zero up to $K_{\max}$, depending on how much energy they lose to internal collisions before escaping the metal.

Question

Wavelengths belonging to Balmer series lying in the range of $450	ext{ nm}$ to $750	ext{ nm}$ were used to eject photoelectrons from a metal surface whose work function is $2.0	ext{ eV}$. Find (in eV) the maximum kinetic energy of the emitted photoelectrons.
(Take $hc = 1242	ext{ eV nm}$.)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are conducting a classic photoelectric effect experiment. You have a metal surface with a known work function, $\Phi = 2.0	ext{ eV}$. This is the minimum energy required to just pull an electron out of the metal's grasp.

Now, instead of shining a simple monochromatic laser on it, you are using light from a hydrogen discharge tube, specifically filtering out the **Balmer series** lines that fall within a wavelength range of $450	ext{ nm}$ to $750	ext{ nm}$.

Our goal is to find the **maximum kinetic energy** ($K_{\max}$) of the electrons that get ejected. According to Einstein's photoelectric equation:
$$K_{\max} = E_{	ext{photon}} - \Phi$$

To maximize $K_{\max}$, we need to find the most energetic photon ($E_{	ext{photon}}$) that is actually present in our filtered light beam.

### The Energy Boundaries

First, let's translate our wavelength filter into an energy filter. We know that the energy of a photon is inversely proportional to its wavelength:
$$E = \frac{hc}{\lambda}$$

Given $hc = 1242	ext{ eV nm}$, we can find the energy boundaries.
For the longest wavelength, $\lambda = 750	ext{ nm}$, the energy is minimum:
$$E_{\min} = \frac{1242}{750} = 1.656	ext{ eV}$$

For the shortest wavelength, $\lambda = 450	ext{ nm}$, the energy is maximum:
$$E_{\max} = \frac{1242}{450} = 2.76	ext{ eV}$$

So, any photon that hits our metal plate must have an energy strictly in the range of **$1.656	ext{ eV}$ to $2.76	ext{ eV}$**.

### The Balmer Series Transitions

Now, we must ask ourselves: *What specific photons are actually being emitted by the hydrogen atoms?*

The Balmer series consists of transitions where an electron falls from a higher energy level ($n > 2$) down to the $n = 2$ level. The energy of an electron in the $n^{	ext{th}}$ orbit is given by Bohr's formula:
$$E_n = -\frac{13.6}{n^2}	ext{ eV}$$

Let's calculate the energy of the $n=2$ state:
$$E_2 = -\frac{13.6}{2^2} = -3.4	ext{ eV}$$

Now, let's test the transitions one by one to see which photons are emitted and if they pass through our filter.

**Transition $n = 3 ightarrow 2$:**
$$E_3 = -\frac{13.6}{3^2} = -1.51	ext{ eV}$$
$$\Delta E_{3 ightarrow 2} = -1.51 - (-3.4) = 1.89	ext{ eV}$$
This photon is within our range ($1.656	ext{ eV} \le 1.89	ext{ eV} \le 2.76	ext{ eV}$).

**Transition $n = 4 ightarrow 2$:**
$$E_4 = -\frac{13.6}{4^2} = -0.85	ext{ eV}$$
$$\Delta E_{4 ightarrow 2} = -0.85 - (-3.4) = 2.55	ext{ eV}$$
This photon is also within our range!

**Transition $n = 5 ightarrow 2$:**
$$E_5 = -\frac{13.6}{5^2} = -0.544	ext{ eV}$$
$$\Delta E_{5 ightarrow 2} = -0.544 - (-3.4) = 2.856	ext{ eV}$$
Ah! This photon has an energy of $2.856	ext{ eV}$, which is greater than our upper limit of $2.76	ext{ eV}$. This means its wavelength is shorter than $450	ext{ nm}$, so it gets blocked by our filter.

### Final Calculation

Out of the photons that actually hit the metal, the most energetic one comes from the $n = 4 ightarrow 2$ transition, carrying **$2.55	ext{ eV}$** of energy.

Now, we bring back Einstein's equation to find the maximum kinetic energy of the ejected electrons:
$$K_{\max} = E_{	ext{photon}} - \Phi$$
$$K_{\max} = 2.55	ext{ eV} - 2.0	ext{ eV}$$
$$K_{\max} = \mathbf{0.55	ext{ eV}}$$

And there we have it! The fastest electrons flying off the metal surface will have a kinetic energy of $0.55	ext{ eV}$.

Enter Numerical Value:

Visualized Solution

$K_{m a x} and λ_{m i n}$

$Energy Range of Incident Light$

$Allowed Energy Range$

$Balmer Series Transitions$

$Evaluating Transition Energies$

$Selecting the Maximum Valid Energy$

$Calculating K_{m a x}$

The Sigma Insight: Photoelectric Effect

Analyzing the Setup

The Energy Boundaries

The Balmer Series Transitions

Final Calculation

Similar Questions

In a photoelectric effect experiment, the threshold wavelength of light is $380 nm$ . If the wavelength of incident light is $260 nm$ , the maximum kinetic energy of emitted electrons will be Given, $E (in eV) = \frac{1237}{λ (in nm)}$

When the wavelength of radiation falling on a metal is changed from $500 nm$ to $200 nm$ , the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to

The work function of a substance is $4.0 eV$ . The longest wavelength of light that can cause photoelectron emission from this substance is approximately

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kmax​ and λmin​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy Range of Incident Light

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Allowed Energy Range

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Balmer Series Transitions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Transition Energies

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Selecting the Maximum Valid Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Kmax​