Question

A sample of excited single electron ions can emit maximum 10 different spectral lines when de-excited to ground level. The maximum energy emitted during transition is 117.5 eV. Now, photons of minimum energy emitted during transition strikes on a metal having work function 1.254 eV emits photoelectrons. The de-Broglie wavelength of emitted photo electron (in Å) is:-

Answer 1

10 Å

Answer 2

Step 1. Determine the Initial Level (n):

A hydrogen-like ion with a single electron produces a maximum number of spectral lines given by

$$N = \frac{n(n-1)}{2}$$

We are told $N = 10$. Thus,

$$\frac{n(n-1)}{2} = 10 \quad \Rightarrow \quad n(n-1) = 20.$$

Trying $n = 5$ gives $5 \times 4 = 20$. So, the ion was excited to the $n=5$ level.

Step 2. Find the Atomic Number (Z):

For a hydrogen-like ion, the energy of a level is

$$E_n = -\frac{Z^2 \cdot 13.6 \text{ eV}}{n^2}.$$

The maximum energy photon is emitted in the transition from $n=5$ to $n=1$, so:

$$\Delta E_{5\to1} = E_1 - E_5 = -\frac{Z^2\cdot 13.6}{1^2} - \left(-\frac{Z^2\cdot 13.6}{25}\right) = Z^2 \cdot 13.6\left(1-\frac{1}{25}\right).$$

Simplify:

$$\Delta E_{5\to1} = Z^2 \cdot 13.6 \cdot \frac{24}{25} = Z^2 \cdot 13.056 \text{ eV}.$$

Given $\Delta E_{5\to1} = 117.5$ eV,

$$Z^2 = \frac{117.5}{13.056} \approx 9 \quad \Rightarrow \quad Z = 3.$$

Step 3. Determine the Minimum Photon Energy:

Among the possible transitions from $n=5$, the smallest energy difference is between adjacent levels, i.e., $5\to4$.

Calculate:

$$E_5 = -\frac{9\cdot 13.6}{25} = -\frac{122.4}{25} \approx -4.896 \text{ eV},$$ $$E_4 = -\frac{9\cdot 13.6}{16} = -\frac{122.4}{16} \approx -7.65 \text{ eV}.$$

Thus,

$$\Delta E_{5\to4} = E_4 - E_5 = (-7.65)-(-4.896) \approx 2.754 \text{ eV}.$$

Step 4. Photoelectric Effect & Kinetic Energy of Photoelectrons:

When a photon of energy $\Delta E_{5\to4} \approx 2.754$ eV strikes a metal with work function $\phi = 1.254$ eV, the kinetic energy $K$ of the emitted photoelectron is:

$$K = \text{Photon Energy} - \phi = 2.754 - 1.254 = 1.5 \text{ eV}.$$

Step 5. Calculate the de Broglie Wavelength:

The de Broglie wavelength is given by:

$$\lambda = \frac{h}{\sqrt{2mK}}.$$

For electrons, a useful formula (with energy $K$ in eV and wavelength $\lambda$ in Å) is:

$$\lambda (\text{\AA}) \approx \frac{12.27}{\sqrt{K \text{ (eV)}}}.$$

Substitute $K = 1.5$ eV:

$$\lambda \approx \frac{12.27}{\sqrt{1.5}} \approx \frac{12.27}{1.225} \approx 10 \text{ \AA}.$$

Final Answer: The de Broglie wavelength of the emitted photoelectrons is approximately 10 Å.