the atomic masses of He and Ne are 4 and 20 a.m.u respectively.the value of debrogile wavelength of He gas at -73 celsius is M times that of de brogile wavelength of Ne at 727 celsius.M is

To find the value of M, which represents the ratio of the de Broglie wavelengths of helium (He) gas at -73°C to neon (Ne) gas at 727°C, we first need to understand how to calculate the de Broglie wavelength and how temperature affects the behavior of gas particles. The de Broglie wavelength is given by the formula:

De Broglie Wavelength Formula

The de Broglie wavelength (λ) can be calculated using the equation:

λ = h / p

where:

• h is Planck's constant (approximately 6.626 x 10^-34 J·s)
• p is the momentum of the particle, which can be expressed as p = mv, where m is the mass and v is the velocity.

Understanding the Relationship Between Temperature and Velocity

The velocity of gas particles is influenced by temperature. According to the kinetic theory of gases, the average kinetic energy of gas particles is directly proportional to the absolute temperature (in Kelvin). The formula for average kinetic energy (KE) is:

KE = (3/2)kT

where k is the Boltzmann constant and T is the temperature in Kelvin. From this, we can derive that:

v = √(3kT/m)

Calculating the Wavelengths

Now, let's calculate the de Broglie wavelengths for both helium and neon. First, we need to convert the temperatures from Celsius to Kelvin:

• -73°C = 200 K
• 727°C = 1000 K

Next, we can substitute the values into the velocity equation for both gases:

• For He (mass = 4 a.m.u = 4 x 1.66 x 10^-27 kg):
• v_He = √(3k(200)/m_He)
• For Ne (mass = 20 a.m.u = 20 x 1.66 x 10^-27 kg):
• v_Ne = √(3k(1000)/m_Ne)

Finding the Ratio of Wavelengths

Now, we can find the ratio of the de Broglie wavelengths:

λ_He / λ_Ne = (h / p_He) / (h / p_Ne) = p_Ne / p_He

Substituting the momentum values:

p_He = m_He * v_He and p_Ne = m_Ne * v_Ne

Thus, the ratio becomes:

λ_He / λ_Ne = (m_Ne * v_Ne) / (m_He * v_He)

Substituting Values

Now we substitute the expressions for velocity:

λ_He / λ_Ne = (m_Ne * √(3k(1000)/m_Ne)) / (m_He * √(3k(200)/m_He))

After simplifying, we find:

λ_He / λ_Ne = √(m_Ne / m_He) * √(1000/200)

Calculating M

Given that:

• m_He = 4 a.m.u
• m_Ne = 20 a.m.u

We can now calculate:

M = λ_He / λ_Ne = √(20/4) * √(1000/200) = √5 * √5 = 5

Therefore, the value of M is 5. This means that the de Broglie wavelength of helium gas at -73°C is 5 times that of neon gas at 727°C.