When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is ________ $$\mathring{A}$$ (angstrom). (Round off to the Nearest Integer).
[Use: $$\sqrt{3} = 1.73$$, $$h = 6.63 \times 10^{-34}$$ Js; $$m_e = 9.1 \times 10^{-31}$$ kg; $$c = 3.0 \times 10^{8}$$ ms$$^{-1}$$; $$1$$ eV $$= 1.6 \times 10^{-19}$$ J]

We are given that light of wavelength $$\lambda = 248$$ nm falls on a metal with threshold energy (work function) $$\phi = 3.0$$ eV. We need to find the de-Broglie wavelength of the emitted electrons.

First, we calculate the energy of the incident photon. Using $$E = \frac{hc}{\lambda}$$, we get $$E = \frac{6.63 \times 10^{-34} \times 3.0 \times 10^{8}}{248 \times 10^{-9}}$$. The numerator is $$6.63 \times 3.0 \times 10^{-34+8} = 19.89 \times 10^{-26}$$ J m. Dividing by $$248 \times 10^{-9}$$ m gives $$E = \frac{19.89 \times 10^{-26}}{248 \times 10^{-9}} = 0.08020 \times 10^{-17} = 8.02 \times 10^{-19}$$ J.

Converting this to electron volts: $$E = \frac{8.02 \times 10^{-19}}{1.6 \times 10^{-19}} = 5.01 \approx 5.0$$ eV.

By Einstein's photoelectric equation, the maximum kinetic energy of the emitted electrons is $$KE = E - \phi = 5.0 - 3.0 = 2.0$$ eV. Converting to joules: $$KE = 2.0 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19}$$ J.

Now, the de-Broglie wavelength is given by $$\lambda_{dB} = \frac{h}{p}$$, where $$p = m_e v$$ is the momentum of the electron. Since $$KE = \frac{p^2}{2m_e}$$, we have $$p = \sqrt{2 \times m_e \times KE}$$.

Substituting the values: $$p = \sqrt{2 \times 9.1 \times 10^{-31} \times 3.2 \times 10^{-19}}$$. Computing the product inside the square root: $$2 \times 9.1 \times 3.2 = 58.24$$, so $$p = \sqrt{58.24 \times 10^{-50}}$$. We can write this as $$p = \sqrt{58.24} \times 10^{-25}$$.

To evaluate $$\sqrt{58.24}$$, we note that $$58.24 = 4 \times 14.56 = 4 \times 3 \times 4.853$$. Alternatively, $$7.6^2 = 57.76$$ and $$7.63^2 \approx 58.22$$. So $$\sqrt{58.24} \approx 7.63$$. Therefore $$p \approx 7.63 \times 10^{-25}$$ kg m s$$^{-1}$$.

The de-Broglie wavelength is $$\lambda_{dB} = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{7.63 \times 10^{-25}} = \frac{6.63}{7.63} \times 10^{-34+25} = 0.8689 \times 10^{-9}$$ m $$= 8.689 \times 10^{-10}$$ m $$= 8.689$$ angstrom.

Rounding off to the nearest integer, the de-Broglie wavelength of the emitted electrons is $$\mathbf{9}$$ angstrom.