I found an answer from chemistry.stackexchange.com

What is the maximum number of emission lines when the excited ...

Mar 17, 2019 **...** Electron doesn't get stuck forever on any of the **levels** with **n**>1. ... excited atom,
but an ensemble of many and many excited **hydrogen atoms**. ... a cascade of
quantized steps of energy loss, say, 6→5→1 or 6→**4**→2→1. ... To **determine N**,
you need to sum the states, as Soumik Das rightfully commented:.

For more information, see What is the maximum number of emission lines when the excited ...

I found an answer from imagine.gsfc.nasa.gov

Solution for Student Worksheet: Energy **Levels** in the **Atom**

May 7, 2015 **...** The equation for **determining** the energy of any state (the nth) is as follows: ... of
the **photon** released when an electron around a **hydrogen atom** ... in the energies
of the fourth (**n**=**4**) and second (n=2) **levels** ... Use the relationship between a
**photon's** energy and its **wavelength** to **calculate** the **wavelength** of ...

For more information, see Solution for Student Worksheet: Energy **Levels** in the **Atom**

I found an answer from www.britannica.com

Photochemical reaction | chemical reaction | Britannica

Ozone protects **Earth's** surface from intense, deep ultraviolet (UV) irradiation, ...
oxygen (O2) into individual oxygen **atoms**, followed by subsequent reaction of ...
not the energy content (i.e., the **wavelength**, colour, or **frequency**) of the radiation.
... number of **absorbed photons** per second, and not their energy, that **determine**
...

For more information, see Photochemical reaction | chemical reaction | Britannica

I found an answer from opentextbc.ca

Bohr's Model of the **Hydrogen Atom** – University **Physics** Volume 3

Explain the **difference between** the absorption **spectrum** and the **emission**
**spectrum** of ... The **emission spectrum** of **atomic hydrogen**: The **spectral** positions
of **emission** lines ... An empirical **formula** to describe the positions (**wavelengths**) \
lambda ... is the energy of either an **emitted** or an absorbed **photon** with
**frequency** f.

For more information, see Bohr's Model of the **Hydrogen Atom** – University **Physics** Volume 3

Given that

Energy levels(quantum number) n = 1,2,3,4

Step 1: Get an expression for wavelength of electron using the Rydberg formula

Rydberg formula

[math]h\upsilon_{12}=\frac{me^4}{8\epsilon_0^2h^2}[\frac{1}{n_1^2}-\frac{1}{n_2^2}][/math]

where, Rydberg constant R = \frac{m e^4}{8 \epsilon_0^2 h^2} = 13.6 eV = 21.76 *10^{-19} J

[math]h\upsilon_{12}=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}][/math]

[math]h\upsilon_{12}=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}][/math]

[math]\frac{hc}{\lambda_{12}}=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}][/math] \because \upsilon = \frac{c}{\lambda}

[math]\lambda_{12}=\frac{hc}{R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]}[/math]

Plugging the known values in the above equation

Higher energy level n_2=4, Lower energy level n_1=1, planck constan h = 6.64 * 10^{-34} , speed of light c = 3* 10^{8}

[math] \lambda_{12} = \frac{6.64 * 10^{-34} * 3 * 10^{8}}{21.76 * 10^{-19} [\frac{1}{1^2} - \frac{1}{4^2}]} [/math]

[math] \lambda_{12} = \frac{1.992*10^{-34}}{21.76 * 10^{-19} [\frac{16 - 1}{16}]} [/math]

\lambda_{12} = 9.1544 * 10^{-8} * \frac{16}{15}

\lambda_{12} = 9.76 * 10^{-8}

Hence, wavelength of the photon \lambda_{12}=9.76*10^{-8} m

Step 2: Calculating the frequency of the photon

Frequency \upsilon = \frac{c}{\lambda_{12}}

\upsilon = \frac{3* 10^8}{9.76 * 10^{-8}}

\upsilon = 3.1*10^{15} Hz

Hence, frequency of the photon \upsilon = 3.1*10^{15} Hz