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

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

Correct answer is option C

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

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

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

= 1.45 $\times$ 10⁹ Hz

Correct answer is option D

As we know, $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mK}}$

If linear momenta of two photons are equal, then their wavelengths is also equal.

Also, if the wavelength is decreased, then the momentum and energy of photon will increase.

Hence, option (d) is correct.

Correct answer is option A

Given, wavelength of x ray ($\lambda$₁) = 1 nm

And wavelength of visible light ($\lambda$₂) = 500 nm

we know, Power$(P)=\frac{nhc}{\lambda }$

As P = constant and h, c also constant

So, $\frac{n}{\lambda }=constant$

$\Rightarrow$ $\frac{n_1}{n_2}=\frac{{\lambda }_1}{{\lambda }_2}=\frac{1nm}{500nm}=\frac{1}{500}$

Correct answer is option A

$2\times {10}^{-3}=\frac{hc}{\lambda }\frac{dn}{dt}$

$\frac{dn}{dt}=\frac{2\times {10}^{-3}\lambda }{hc}$

$=\frac{2\times {10}^{-3}\times 500\times {10}^{-9}}{6.6\times {10}^{-34}\times 3\times {10}^8}$

$=\frac{1000}{6.6\times 3}\times 10=5\times {10}^{15}$

Correct answer is option B

The minimum wavelength of emitted photons is

$\lambda$ = $\frac{1240}{5.6-0.7}nm$ = 250 nm

Correct answer is option A

The power of the given laser light is expressed as

$P=\frac{nhc}{\lambda t}$

from which the number of photons per second is given by

$n=P\left(\frac{\lambda t}{hc}\right)=(5\times {10}^2)\times \left[\frac{(660\times {10}^{-9})(60\times {10}^{-3})}{(6.6\times {10}^{-34})(3\times {10}^8)}\right]$

$=100\times {10}^{18}={10}^{20}$

Correct answer is option A

Power, $P=\frac{nhv}{t}$

$\Rightarrow v=\frac{P\times t}{nh}$

$=\frac{4\times {10}^3\times 1}{{10}^{20}\times 6.63\times {10}^{-34}}=6\times {10}^{16}Hz$

Correct answer is option A

Energy of a photon of frequency $v$ is given by $E=hv.$

Also, $E=m{c}^2,m{c}^2=hv$

$\Rightarrow mc=\frac{hv}{C}\Rightarrow p=\frac{hv}{c}$

Correct answer is option A

$I\propto \frac{1}{r^2};\frac{I_1}{I_2}={\left(\frac{r_2}{r_1}\right)}^2=\frac{1}{4}$

$I_2\to 4$ times $I_1$

When intensity becomes 4 times, no. of photoelectrons emitted would increase by $4$ times, since number of electrons emitted per second is directly proportional to intensity.