Given, ($$\tau$$_1/2)_x = ($$\tau$$)_y

Here, $$\tau$$_1/2 = half-life period of radioactive element and $$\tau$$ = mean life period of radioactive element.

As we know the expression,

Half-life of the radioactive element x,

$${\tau _{1/2}} = {{\ln (2)} \over {{\lambda _x}}}$$

Mean life of the radioactive element y,

$$\tau = {1 \over {{\lambda _y}}}$$

Substituting the values in Eq. (i), we get

$${{\ln 2} \over {{\lambda _x}}} = {1 \over {{\lambda _y}}} \Rightarrow {\lambda _x} = 0.693{\lambda _y}$$

Initially they have same number of atoms,

$${N_x} = {N_y} = {N_0}$$

As we know that,

Activity, $$A = \lambda N$$

As, $${\lambda _x} < {\lambda _y} \Rightarrow {A_x} < {A_y}$$

Therefore, y will decay faster than x.

Given, Planck's constant, h = 6.6 $$\times$$ 10^$$-$$34 Js

Boltzmann constant, k_B = 1.38 $$\times$$ 10^$$-$$23 J/K

Mass of an electron, m_e = 9 $$\times$$ 10^$$-$$31 kg

Temperature of an ideal gas, T = 300 K

As we know that, de-Broglie wavelength,

$$\lambda = {h \over {mv}} = {h \over {\sqrt {2mE} }}$$ .... (i)

Here, E is the kinetic energy,

$$E = {{3{K_B}T} \over 2}$$

Substituting value of E in Eq. (i), we get

$$\lambda = {h \over {\sqrt {3m{K_B}T} }}$$

Substituting the given values in the above equation, we get

$$\lambda = {{6.6 \times {{10}^{ - 34}}} \over {\sqrt {3 \times 9 \times {{10}^{ - 31}} \times 1.38 \times {{10}^{ - 23}} \times 300} }}$$

= 6.26 nm

$$\therefore$$ The corresponding de-Broglie wavelength of an electron approximately at 300 K is 6.26 nm.

$$\Delta x\,.\,\Delta p \ge {h \over {4\pi }}$$

$$\Delta x = {h \over {4\pi m\Delta v}}$$

$$v = \sqrt {{{3KT} \over m}} $$

$${{\Delta {x_e}} \over {\Delta {x_p}}} = \sqrt {{{{m_p}} \over {{m_e}}}} $$

For every large distance P.E. = 0

& total energy = 2.6 + 0 = 2.6 eV

Finally in first excited state of H atom total energy = $$-$$3.4 eV

Loss in total energy = 2.6 $$-$$ ($$-$$3.4) = 6 eV

It is emitted as photon

$$\lambda = {{1240} \over 6} = 206$$ nm

$$f = {{3 \times {{10}^8}} \over {206 \times {{10}^{ - 9}}}}$$ = 1.45 $$\times$$ 10¹⁵ Hz

= 1.45 $$\times$$ 10⁹ Hz