Class 11th
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New answer posted
7 months agoContributor-Level 10
g_h = g (1-2h/R). g_d = g (1-d/R). Given h=d.
g_h = g (R/ (R+h)² ≈ g (1-2h/R). This differs from the image.
The image has g/ (1+h/R)² = g (1-h/R). This leads to h² + hR - R²=0. h = R (√5-1)/2.
New answer posted
7 months agoContributor-Level 10
ΔE = have = 13.6Z² (1/n? ² - 1/n? ²)eV.
hv = 13.6 (1²) (1/n² - 1/ (n+1)²) = 13.6 (n+1)²-n²)/ (n² (n+1)²)
v = (13.6/h) * (2n+1)/ (n² (n+1)²). For n>>1, v ≈ (13.6/h) (2n/n? ) ∝ 1/n³.
New answer posted
7 months agoContributor-Level 10
σ4πr² + σ4πR² = Q ⇒ σ = Q/ (4π (R²+r²)
V_c = kq? /r + kq? /R = k (σ4πr²)/r + k (σ4πR²)/R = kσ4π (r+R)
= K (Q/ (R²+r²) (R+r)
New answer posted
7 months agoContributor-Level 10
On increasing the temperature, random velocity of molecules increases, therefore mean collision time between the molecules decreases. But the mean free path remains constant as it is product of velocity and time.? (b) and (c) are correct.
New answer posted
7 months agoContributor-Level 10
λ = h/p = h/mv
λp/λe = (m? v? )/ (m? v? ) ⇒ 1.878 * 10? = (9.1*10? ³¹)/ (m? * 5)
m? = 9.1*10? ³¹ / (5 * 1.878 * 10? ) = 0.97 * 10? ²? kg
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