M.231 तर m=D?57श्न 2 (A) खालील कृती पू

1.	M.231 तर m=D?57श्न 2 (A) खालील कृती पूर्ण करा. (3 पैकी 2)V2x +7x+5/2 = 0 हे वर्गसमीकरण अवयव पद्धतीने सोडवण्यासाठी पुढील कृती पूर्ण करा./2x +7x+5/2 = 02x5/2 = 5x2=10+5+22x +[]+[]+5V2 = 0..x(/2x +5) +-2(-2x +5) =02.(/2x+5)(x + 2)= 0/2x+5=30 किंवा + 2 = 0किंवा =--2वर्गसमीकरणाची मुळे आणि -2 ही आहेत.-: x=0मराठी मधे
Answer» Answer: √9π Step-by-step explanation: 13. No, because the probability of the particle existing in a narrow (infinitesimally small) interval at the discontinuity is undefined. 15. No. For an infinite square well, the spacing between energy LEVELS increases with the quantum number n. The smallest energy measured corresponds to the transition from n = 2 to 1, which is three times the ground state energy. The largest energy measured corresponds to a transition from n=∞ to 1, which is infinity. (NOTE: Even PARTICLES with extremely LARGE energies remain BOUND to an infinite square well—they can never “escape”)

M.231 तर m=D?57श्न 2 (A) खालील कृती पूर्ण करा. (3 पैकी 2)V2x +7x+5/2 = 0 हे वर्गसमीकरण अवयव पद्धतीने सोडवण्यासाठी पुढील कृती पूर्ण करा./2x +7x+5/2 = 02x5/2 = 5x2=10+5+22x +[]+[]+5V2 = 0..x(/2x +5) +-2(-2x +5) =02.(/2x+5)(x + 2)= 0/2x+5=30 किंवा + 2 = 0किंवा =--2वर्गसमीकरणाची मुळे आणि -2 ही आहेत.-: x=0मराठी मधे

Answer»

Answer: √9π

Step-by-step explanation:

13. No, because the probability of the particle existing in a narrow (infinitesimally small) interval at the discontinuity is undefined.

15. No. For an infinite square well, the spacing between energy LEVELS increases with the quantum number n. The smallest energy measured corresponds to the transition from n = 2 to 1, which is three times the ground state energy. The largest energy measured corresponds to a transition from n=∞ to 1, which is infinity. (NOTE: Even PARTICLES with extremely LARGE energies remain BOUND to an infinite square well—they can never “escape”)

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