Correct answer is option A

$\begin{array}{rl}& \text{ Energy of one photon }=\frac{\text{ Power }}{\text{ Photon frequency }}\\ \\ & \mathrm{E}=\mathrm{h}v=\frac{15\times {10}^3}{{10}^{16}}\\ \\ & \Rightarrow v=\frac{15\times {10}^{-13}}{6\times {10}^{-34}}=2.5\times {10}^{21}\end{array}$

So gamma Rays.

correct answer is 4125

# Explanation

The energy $E$ of a photon is related to its wavelength $\lambda$ by the formula:

$E=\frac{hc}{\lambda }$

where $h$ is Planck's constant and $c$ is the speed of light. In this problem, we are given that an atom absorbs a photon of wavelength ${\lambda }_1=500\mathrm{nm}$ and emits another photon of wavelength ${\lambda }_2=600\mathrm{nm}$. We can use the formula above to calculate the energy absorbed by the atom:

$\mathrm{Energy}\mathrm{absorbed}=E_1-E_2=\frac{hc}{{\lambda }_1}-\frac{hc}{{\lambda }_2}=hc(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2})$

Substituting the given values for $h$ and $c$, we get:

$\mathrm{Energy}\mathrm{absorbed}=6.6\times {10}^{-34}\mathrm{J}\cdot \mathrm{s}\cdot 3\times {10}^8\mathrm{m}/\mathrm{s}\cdot (\frac{1}{500\times {10}^{-9}\mathrm{m}}-\frac{1}{600\times {10}^{-9}\mathrm{m}})$

Simplifying this expression, we get:

$\mathrm{Energy}\mathrm{absorbed}=6.6\times {10}^{-20}\mathrm{J}$

We need to express this energy in electron volts (eV), which is a more convenient unit for atomic and molecular energies. To do this, we can divide the energy in joules by the charge of an electron:

$\mathrm{Energy}\mathrm{absorbed}\mathrm{in}\mathrm{eV}=\frac{6.6\times {10}^{-20}\mathrm{J}}{1.6\times {10}^{-19}\mathrm{C}/\mathrm{eV}}=0.4125\mathrm{eV}$

Finally, we can express the net energy absorbed in terms of the given value of $n$, as follows:

$n\times {10}^{-4}\mathrm{eV}=0.4125\mathrm{eV}$

Solving for $n$, we get:

$n=\frac{0.4125}{{10}^{-4}}=4125$

Therefore, the value of $n$ is $\boxed{{\displaystyle 4125}}$.

correct answer is 3

# Explanation

When a monochromatic light is incident on hydrogen atoms in the ground state (n = 1), the hydrogen atoms can absorb energy and transition to higher energy levels. When the atoms return to lower energy levels, they emit radiation of different wavelengths corresponding to the energy differences between the energy levels.

The energy levels of the hydrogen atom are given by the formula:

$E_n=-\frac{13.6\mathrm{eV}}{n^2}$

where $E_n$ is the energy of the nth level and $n$ is the principal quantum number.

Since the hydrogen atoms emit radiation of six different wavelengths, there must be six different transitions from the excited states back to lower energy levels.

The six transitions correspond to the following energy level changes:

1. From n = 2 to n = 1
2. From n = 3 to n = 1
3. From n = 3 to n = 2
4. From n = 4 to n = 1
5. From n = 4 to n = 2
6. From n = 4 to n = 3

The highest energy level involved is n = 4. Therefore, the incident light must have a frequency high enough to excite the hydrogen atoms from the ground state (n = 1) to n = 4.

The energy difference between these levels is:

$\mathrm{\Delta }E=E_4-E_1=\frac{13.6\mathrm{eV}}{4^2}-\frac{13.6\mathrm{eV}}{1^2}=-0.85\mathrm{eV}+13.6\mathrm{eV}=12.75\mathrm{eV}$

The frequency of the incident light is related to the energy difference by the equation:

$\mathrm{\Delta }E=h\nu$

where $h$ is the Planck's constant and $\nu$ is the frequency.

Now, we can solve for the frequency:

$\nu =\frac{\mathrm{\Delta }E}{h}=\frac{12.75\mathrm{eV}}{4.25\times {10}^{-15}\mathrm{eVs}}=3\times {10}^{15}\mathrm{Hz}$

So, the value of $x$ is 3.

Correct answer is option B

$L$ = 900 nm

I = 100 W/m²

A = 10^{$-$ 4}

$\Rightarrow$ P = 10 ^{$-$ 2} W

$\Rightarrow$ Number of photons incident per second

$=\frac{{10}^{-2}\lambda }{hc}$

$=\frac{9\times {10}^{-11}\times {10}^2}{6.63\times {10}^{-34}\times 3\times {10}^8}\simeq 4.5\times {10}^{16}$

Correct answer is option A

As we know

$\begin{array}{rl}& I=\frac{E}{At}=\frac{nhv}{At}\Rightarrow \frac{n}{t}=\frac{IA\lambda }{hC}\Rightarrow \frac{n}{t}=\frac{\rho \lambda }{hc}\Rightarrow \frac{n}{t}=\rho \lambda \\ \\ & \Rightarrow {\left(\frac{n}{t}\right)}_2={\left(\frac{n}{t}\right)}_1\times \frac{p_2{\lambda }_2}{p_1\lambda 1}=9\times {10}^{20}\times \frac{P}{P}\times \frac{800}{600}=12\times {10}^{20}\end{array}$