Question

A hydrogen-1 atom in the ground state absorbs a UV photon. As a result, the 1H atom is ionized – it breaks into a proton moving with a speed of 1.15×103 m/s and an electron moving with a speed of 2.11×106 m/s in the opposite direction. Calculate the wavelength of the UV photon. [Note: masses of electron and proton are available in the textbook and on the internet; use masses in kg.]

Answer 1

`\lambda = 390 nm`

Step-by-step explanation:

Here we can use momentum conservation to find the initial wavelength of UV photon

so the equation for momentum conservation is given as

`(h)/(\lambda) = m_1v_1 + m_2v_2`

we will have

`v_1 = 1.15 * 10^3 m/s`

`m_1 = 1.67 * 10^(-27) kg`

`v_2 = -2.11 * 10^6 m/s`

`m_2  = 9.11 * 10^(-31) kg`

so we will have

`(h)/(\lambda) = (1.67 * 10^(-27))(1.15 * 10^3) - (9.11 * 10^(-31))(2.11 * 10^6)`

`(h)/(\lambda) = -1.71 * 10^(-27)`

`\lambda = 390 nm`